RD Sharma Class 12 Exercise 18.30 Indefinite Integrals Solutions Maths

RD Sharma Class 12 Exercise 18.30 Indefinite Integrals Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 10:32 AM IST

The RD Sharma class 12 chapter 18 exercise 18.30 answers are trusted by the students on a large scale. Mathematics is a subject where many students lose marks due to lack of good practice. Therefore, it is essential for every student to own the RD Sharma Class 12 Exercise 18.30 reference book to perform well in their public exams. The 18th chapter of the class 12 mathematics syllabus consists of 31 exercises, ex 18.1 to ex 18.31. This counts to hundreds of questions that is present in this single chapter. RD Sharma solutions When it comes to ex 18.30, the difficulty of the questions would be higher as the exercises reach the end of the chapter.

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RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.30

Indefinite Integrals exercise 18.30 question 1

Answer:
$\frac{1}{3} \log |x+1|+\frac{5}{3} \log |x-2|+C$
Hint:
To solve the given integration, we use partial fraction method
Given:
$\int \frac{2 x+1}{(x+1)(x-2)} d x$
Explanation:
Let
$I=\int \frac{2 x+1}{(x+1)(x-2)} d x$
Expressing the function in terms of partial fraction as following
$\left.\frac{2 x+1}{(x+1)(x-2)}=\frac{A}{(x+1)}+\frac{B}{(x-2)} \quad \text { [Using } \frac{N(x)}{(a x+b)(c x+d)}=\frac{A}{a x+b}+\frac{B}{c x+d}\right]$
$\begin{aligned} &\frac{2 x+1}{(x+1)(x-2)}=\frac{A(x-2)+B(x+1)}{(x+1)(x-2)} \\ &2 x+1=A(x-2)+B(x+1) \\ &2 x+1=(A+B) x+(-2 A+B) \end{aligned}$
On comparing the coefficient of x and constant terms we get
$\begin{gathered} 2=A+B \\ \end{gathered}$ (1)
$\begin{aligned} -2 A+B=1 \end{aligned}$ (2)
Now, subtract equation (2) from (1)
$\begin{gathered} A+B=2- \\ \frac{-2 A+B=1}{3 A=1} \\ A=\frac{1}{3} \end{gathered}$
Putting the value of A in equation (1)
$\begin{aligned} &\frac{1}{3}+B=2 \\ &B=2-\frac{1}{3} \\ &B=\frac{5}{3} \end{aligned}$
Now
$\begin{aligned} &\frac{2 x+1}{(x+1)(x-2)}=\frac{1}{3(x+1)}+\frac{5}{3(x-2)} \\ &I=\int\left(\frac{1}{3(x+1)}+\frac{5}{3(x-2)}\right) d x \\ &=\int \frac{1}{3(x+1)} d x+\int \frac{5}{3(x-2)} d x \end{aligned}$
$\left.=\frac{1}{3} \log |x+1| d x+\frac{5}{3} \log |x-2|+C \quad \text { [Using } \int \frac{1}{a x+b} d x=\frac{1}{a} \log |a x+b|+C\right]$

Indefinite Integrals exercise 18.30 question 3

Answer:
$x+\log \left|\frac{x-2}{x+3}\right|+C$
Hint:
To solve the given integration, we use partial fraction method
Given:
$\int \frac{x^{2}+x-1}{x^{2}+x-6} d x$
Explanation:
Let
$I=\int \frac{x^{2}+x-1}{x^{2}+x-6} d x$
$I=\int \frac{\left(x^{2}+x-6\right)+5}{x^{2}+x-6} d x$ [Adding and subtracting 5 in numerator]
$\begin{aligned} &\int\left(1+\frac{5}{x^{2}+x-6}\right) d x\\ &I=\int d x+\int \frac{5}{x^{2}+x-6} d x\\ &I=x+\int \frac{5}{x^{2}+x-6} d x \end{aligned}$ (1)
Let
$I_{1}=\int \frac{5}{(x+3)(x-2)} d x \quad\left[x^{2}+x-6=(x+3)(x-2)\right]$
Now express the function in terms of partial fraction
$\frac{5}{(x+3)(x-2)}=\frac{A}{x+3}+\frac{B}{x-2} \quad\left[\frac{N(x)}{(a x+b)(c x+d)}=\frac{A}{a x+b}+\frac{B}{c x+d}\right]$
$\begin{aligned} &\frac{5}{(x+3)(x-2)}=\frac{A(x-2)+B(x+3)}{(x+3)(x-2)} \\ &5=A(x-2)+B(x+3) \\ &5=(A+B) x+(-2 A+3 B) \end{aligned}$
On comparing the coefficient we get
$\begin{aligned} &A+B=0\quad\quad\quad(2)\\ &-2 A+3 B=5\quad\quad(3)\\ &A=-B \end{aligned}$

Equation (3)

$\begin{aligned} &2 B+3 B=5 \\ &5 B=5 \\ &B=1 \\ &A=-1 \end{aligned}$
Now
$\begin{aligned} &I_{1}=\int\left(\frac{-1}{x+3}+\frac{1}{x-2}\right) d x \\ &I_{1}=-\int \frac{1}{x+3} d x+\int \frac{1}{x-2} d x \end{aligned}$
$\begin{aligned} &I_{1}=-\log |x+3|+\log |x-2|+C \quad\left[\int \frac{1}{a x+b} d x=\frac{1}{a} \log |a x+b|\right] \\ &I_{1}=\log \left|\frac{x-2}{x+3}\right|+C \end{aligned}$
Putting the value of $I_{1}$ in equation (1)
$I=x+\log \left|\frac{x-2}{x+3}\right|+C$

Indefinite Integrals exercise 18.30 question 2

Answer:
$\frac{1}{8} \log \left|\frac{x(x-4)}{(x-2)^{2}}\right|+C$
Hint:
To solve the given integration, we use partial fraction method
Given:
$\int \frac{1}{x(x-2)(x-4)} d x$
Explanation:
Let $I=\int \frac{1}{x(x-2)(x-4)} d x$
Now express the functions in terms of partial fraction
$\frac{1}{x(x-2)(x-4)}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{x-4} \quad\left[\because \frac{N(x)}{(a x-b)(c x-d)}=\frac{A}{a x-b}+\frac{B}{c x-d}\right]$
$\begin{aligned} &\frac{1}{x(x-2)(x-4)}=\frac{A(x-2)(x-4)+B x(x-4)+C x(x-2)}{x(x-2)(x-4)} \\ &1=A\left(x^{2}-6 x+8\right)+B\left(x^{2}-4 x\right)+C\left(x^{2}-2 x\right) \\ &1=x^{2}(A+B+C)+x(-6 A-4 B-2 C)+8 A \end{aligned}$
On comparing coefficient, we get
$\begin{aligned} &A+B+C=0\quad\quad\quad\quad(1)\\ &-6 A-4 B-2 C=0\quad\quad\quad(2)\\ &8 A=1 \quad\quad\quad\quad(3)\\ &A=\frac{1}{8} \end{aligned}$
Now equation (2)
$\begin{aligned} &\Rightarrow \frac{-6}{8}-4 B-2 C=0 \\ &\Rightarrow 2 B+C=\frac{-3}{8} \end{aligned}$ (4)
Equation (1)
$\begin{aligned} &\Rightarrow \frac{1}{8}+B+C=0 \\ &\Rightarrow B+C=\frac{-1}{8} \end{aligned}$ (5)
Subtract equation (5) from equation (4)
$\begin{gathered} 2 B+C=\frac{-3}{8}- \\ B+C=\frac{-1}{8} \\ \hline B=\frac{-2}{8} \end{gathered}$
Now equation (5)
$\begin{aligned} &\Rightarrow \frac{-2}{8}+C=\frac{-1}{8} \\ &C=\frac{-1}{8}+\frac{2}{8} \\ &C=\frac{1}{8} \end{aligned}$
Now
$\begin{aligned} &\frac{1}{x(x-2)(x-4)}=\left(\frac{1}{8 x}\right)+\frac{(-2)}{8(x-2)}+\frac{1}{8(x-4)} \\ &I=\int\left(\frac{1}{8 x}-\frac{2}{8(x-2)}+\frac{1}{8(x-4)}\right) d x \\ &I=\frac{1}{8} \int \frac{1}{x} d x-\frac{2}{8} \int \frac{1}{x-2} d x+\frac{1}{8} \int \frac{d x}{x-4} \end{aligned}$
$\begin{aligned} &I=\frac{1}{8} \log |x|-\frac{2}{8} \log |x-2|+\frac{1}{8} \log |x-4|+C \quad\left[\int \frac{1}{a x-b} d x=\frac{1}{a} \log |a x-b|\right] \\ &I=\frac{1}{8} \log \left|\frac{x(x-4)}{(x-2)^{2}}\right|+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 4

Answer:
$-x+2 \log |x-1|+3 \log |x+2|+C$
Hint:
To solve the given integration, we use partial fraction method
Given:
$\int \frac{3+4 x-x^{2}}{(x+2)(x-1)} d x$
Explanation:
Let
$I=\int \frac{3+4 x-x^{2}}{(x+2)(x-1)} d x$
$I=\int \frac{3+4 x-\left(x^{2}+x-2\right)+x-2}{x^{2}+x-2} d x$ [Adding and subtract $\left ( x-2 \right )$ in numerator]
$\begin{aligned} &I=\int \frac{1+5 x-\left(x^{2}+x-2\right)}{x^{2}+x-2} d x\\ &I=\int\left(\frac{1+5 x}{x^{2}+x-2}-1\right) d x\\ &I=\int \frac{1+5 x}{(x-1)(x+2)} d x-\int d x\\ &I=\int \frac{1+5 x}{(x-1)(x+2)} d x-x \quad \quad \quad(1) \end{aligned}$
Let
$\begin{aligned} &I_{1}=\int \frac{1+5 x}{(x-1)(x+2)} d x \\ &\frac{1+5 x}{(x-1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+2} \quad\left[\frac{N(x)}{(a x+b)(c x+d)}=\frac{A}{a x+b}+\frac{B}{c x+d}\right] \end{aligned}$

$\begin{aligned} &\frac{1+5 x}{(x-1)(x+2)}=\frac{A(x+2)+B(x-1)}{(x-1)(x+2)} \\ &(1+5 x)=(A+B) x+(2 A-B) \end{aligned}$
Comparing the corresponding coefficient
$\begin{aligned} &A+B=5\quad \text { (2) }\\ &2 A-B=1 \quad \text { (3) } \end{aligned}$

Adding equation (2) and (3)

$\begin{aligned} &A+B=5+ \\ &\frac{2 A-B=1}{3 A=6} \\ &A=2 \end{aligned}$
Now equation (2)
$\begin{aligned} &2+B=5 \\ &B=3 \end{aligned}$
Now
$\begin{aligned} &I_{1}=\int\left(\frac{2}{x-1}+\frac{3}{x+2}\right) d x \\ &I_{1}=2 \int \frac{1}{x-1} d x+3 \int \frac{1}{x+2} d x \\ &I_{1}=2 \log |x-1|+3 \log |x+2|+C \end{aligned}$
Now putting the value of $I_{1}$ in equation (1) and we get
$I=-x+2 \log |x-1|+3 \log |x+2|+C$

Indefinite Integrals exercise 18.30 question 5

Answer:
$x+\log \left|\frac{x-1}{x+1}\right|+C$
Hint:
To solve the given integration, first we write the function in simple form and then apply the formula of integration
Given:
$\int \frac{x^{2}+1}{x^{2}-1} d x$
Explanation:
$\begin{aligned} &I=\int \frac{x^{2}+1}{x^{2}-1} d x \\ &I=\int \frac{x^{2}-1+2}{x^{2}-1} d x \\ &I=\int\left(1+\frac{2}{x^{2}-1}\right) d x \\ &I=\int d x+\int \frac{2}{x^{2}-1} d x \end{aligned}$
$\begin{aligned} &I=x+2\left(\frac{1}{2} \log \left|\frac{x-1}{x+1}\right|\right)+C \quad\left[\int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|\right] \\ &I=x+\log \left|\frac{x-1}{x+1}\right|+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 6

Answer:
$\frac{1}{2} \log |x-1|-4 \log |x-2|+\frac{9}{2} \log |x-3|+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{x^{2}}{(x-1)(x-2)(x-3)} d x$
Explanation:
$\begin{aligned} &I=\int \frac{x^{2}}{(x-1)(x-2)(x-3)} d x \\ &\frac{x^{2}}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3} \quad\left[\frac{p x^{2}+q x+c}{(x-a)(x-b)(x-c)}=\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}\right] \end{aligned}$
$\begin{aligned} &\frac{x^{2}}{(x-1)(x-2)(x-3)}=\frac{A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)}{(x-1)(x-2)(x-3)} \\ &x^{2}=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2) \end{aligned}$
$\begin{aligned} &\text { At } x=2 \\ &4=0+B(1)(-1)+0 \\ &4=-B \\ &B=-4 \end{aligned}$
$\begin{aligned} &\text { At } x=3 \\ &9=0+0+C(2)(1) \\ &9=2 C \\ &C=\frac{9}{2} \end{aligned}$
$\begin{aligned} &\text { At } x=1 \\ &1=A(-1)(-2)+0+0 \\ &2 A=1 \\ &A=\frac{1}{2} \end{aligned}$
Now
$\begin{aligned} &\frac{x^{2}}{(x-1)(x-2)(x-3)}=\frac{1}{2(x-1)}+\frac{(-4)}{x-2}+\frac{9}{2(x-3)} \\ &I=\int\left[\frac{1}{2(x-1)}-\frac{4}{x-2}+\frac{9}{2(x-3)}\right] d x \\ &I=\frac{1}{2} \int \frac{1}{x-1} d x-4 \int \frac{1}{x-2} d x+\frac{9}{2} \int \frac{1}{x-3} d x \\ &I=\frac{1}{2} \log |x-1|-4 \log |x-2|+\frac{9}{2}|x-3|+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 7

Answer:
$\frac{5}{3} \log |x+1|+\frac{5}{6} \log |x-2|-\frac{5}{2} \log |x+2|+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{5 x}{(x+1)\left(x^{2}-4\right)} d x$
Explanation:
Let
$\begin{aligned} &I=\int \frac{5 x}{(x+1)\left(x^{2}-4\right)} d x \\ &\left.I=\int \frac{5 x}{(x+1)(x-2)(x+2)} d x \ldots \text { (applying the } \text { formula } a^{2}-b^{2}\right) \end{aligned}$
$\begin{aligned} &\frac{5 x}{(x+1)(x-2)(x+2)}=\frac{A}{x+1}+\frac{B}{x-2}+\frac{C}{x+2} \\\\ &\frac{5 x}{(x+1)(x-2)(x+2)}=\frac{A(x-2)(x+2)+B(x+1)(x+2)+C(x+1)(x-2)}{(x+1)(x-2)(x+2)} \\\\ &5 x=A(x-2)(x+2)+B(x+1)(x+2)+C(x+1)(x-2) \end{aligned}$
$\begin{aligned} &\text { At } x=2 \\ &5(2)=0+B(3)(4)+0 \\ &10=12 B \\ &B=\frac{10}{12} \\ &B=\frac{5}{6} \end{aligned}$
$\begin{aligned} &\text { At } x=-2 \\ &5(-2)=0+0+C(-1)(-4) \\ &-10=4 C \\ &C=\frac{-10}{4} \\ &\text { At } x=-1 \end{aligned}$
$\begin{aligned} &5(-1)=A(-3)(1)+0+0 \\ &-5=-3 A \\ &\begin{array}{l} A=\frac{5}{3} \\\\ \end{array} \end{aligned}$
$\frac{5 x}{(x+1)(x-2)(x+2)}=\frac{5}{3(x+1)}+\frac{5}{6(x-2)}-\frac{5}{2(x+2)}$
$\begin{aligned} &I=\int\left(\frac{5}{3(x+1)}+\frac{5}{6(x-2)}-\frac{5}{2(x+2)}\right] d x \\ &I=\frac{5}{3} \int \frac{1}{x+1} d x+\frac{5}{6} \int \frac{1}{x-2} d x-\frac{5}{2} \int \frac{1}{x+2} d x \\ &I=\frac{5}{3} \log |x+1|+\frac{5}{6} \log |x-2|-\frac{5}{2} \log |x+2|+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 8

Answer:
$\log \left|\frac{x^{2}-1}{x}\right|+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{x^{2}+1}{x\left(x^{2}-1\right)} d x$
Explanation:
Let
$\begin{aligned} &I=\int \frac{x^{2}+1}{x\left(x^{2}-1\right)} d x \\ &I=\int \frac{\left(x^{2}-1\right)+2}{x\left(x^{2}-1\right)} d x \end{aligned}$ [Add and subtract 1]
$\begin{aligned} &I=\int\left(\frac{1}{x}+\frac{2}{x\left(x^{2}-1\right)}\right) d x \\ &I=\int \frac{1}{x} d x+2 \int \frac{1}{x(x-1)(x+1)} d x \end{aligned}$ …(applying the formula $a^{2}-b^{2}$ )
$I=\log |x|+2 I_{1}$ (1)
Where
$\begin{aligned} &I_{1}=\int \frac{1}{x(x-1)(x+1)} d x \\ &\frac{1}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1} \\ &\frac{1}{x(x-1)(x+1)}=\frac{A(x-1)(x+1)+B x(x+1)+C x(x-1)}{x(x-1)(x+1)} \\ &1=A(x-1)(x+1)+B(x)(x+1)+C(x)(x-1) \end{aligned}$
$\begin{aligned} &\text { At } x=0 \\ &1=A(-1)(1)+0+0 \\ &A=-1 \end{aligned}$
$\begin{aligned} &\text { At } x=1 \\ &1=0+B(1)(2)+0 \\ &1=2 B \\ &B=\frac{1}{2} \end{aligned}$
$\begin{aligned} &\text { At } x=-1 \\ &1=0+0+C(-1)(-2) \\ &1=2 C \\ &C=\frac{1}{2} \end{aligned}$
$\begin{aligned} &\frac{1}{(x-1)(x+1) x}=\frac{-1}{x}+\frac{1}{2(x-1)}+\frac{1}{2(x+1)} \\ &I_{1}=\int\left[\frac{-1}{x}+\frac{1}{2(x-1)}+\frac{1}{2(x+1)}\right] d x \\ &I_{1}=-\int \frac{1}{x} d x+\frac{1}{2} \int \frac{1}{x-1} d x+\frac{1}{2} \int \frac{1}{x+1} d x \\ &I_{1}=-\log |x|+\frac{1}{2} \log |x-1|+\frac{1}{2} \log |x+1|+C \end{aligned}$
Equation (1)
$I=\log |x|-2 \log |x|+\log |x-1|+\log |x+1|+C$
$\begin{aligned} &I=-\log |x|+\log |(x-1)(x+1)|+C \\ &I=\log \left|\frac{x^{2}-1}{x}\right|+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 9

Answer:
$\frac{-1}{10} \log |x-1|+\frac{5}{2} \log |x+1|-\frac{12}{5} \log |2 x+3|+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{2 x-3}{(x-1)(x+1)(2 x+3)}$
Explanation:
Let
$I=\int \frac{2 x-3}{(x-1)(x+1)(2 x+3)}$ …(applying the formula $a^{2}-b^{2}$ )
$\frac{2 x-3}{\left(x^{2}-1\right)(2 x+3)}=\frac{2 x-3}{(x-1)(x+1)(2 x+3)}$
$\begin{aligned} &\frac{2 x-3}{(x-1)(x+1)(2 x+3)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{(2 x+3)} \\ &\frac{2 x-3}{(x-1)(x+1)(2 x+3)}=\frac{A(x+1)(2 x+3)+B(x-1)(2 x+3)+C(x-1)(x+1)}{(x-1)(x+1)(2 x+3)} \\ &2 x-3=A(x+1)(2 x+3)+B(x-1)(2 x+3)+C(x-1)(x+1) \end{aligned}$
$\begin{aligned} &\text { At } x=1 \\ &2 \times 1-3=A(2)(5)+B(0)+C(0) \\ &-1=10 \mathrm{~A} \\ &A=\frac{-1}{10} \end{aligned}$

$\begin{aligned} &\text { At } x=-1 \\ &2 \times-1-3=A(0)+B(-2)(1)+0 \\ &-5=-2 B \\ &B=\frac{5}{2} \end{aligned}$
$\begin{aligned} &\text { At } x=\frac{-3}{2} \\ &2 \times\left(\frac{-3}{2}\right)-3=A(0)+B(0)+C\left(\frac{-3}{2}+1\right)\left(\frac{-3}{2}-1\right) \\ &-6=C\left(\frac{5}{4}\right) \\ &C=\frac{-24}{5} \end{aligned}$
$\frac{2 x-3}{(x-1)(x+1)(2 x+3)}=\frac{-1}{10(x-1)}+\frac{5}{2(x+1)}-\frac{24}{5(2 x+3)}$

$\begin{aligned} &I=\frac{-1}{10} \int \frac{1}{x-1} x+\frac{5}{2} \int \frac{1}{x+1} d x-\frac{24}{5} \int \frac{1}{2 x+3} d x \\ &I=\frac{-1}{10} \log |x-1|+\frac{5}{2} \log |x+1|-\frac{12}{5} \log |2 x+3|+C \\ &I=\frac{-1}{10} \log |x-1|+\frac{5}{2} \log |x+1|-\frac{12}{5} \log |2 x+3|+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 10

Answer:
$x+\frac{1}{2} \log |x-1|-8 \log |x-2|+\frac{27}{2} \log |x-3|+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{x^{3}}{(x-1)(x-2)(2 x+3)} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{x^{3}}{(x-1)(x-2)(2 x+3)} d x \\ &I=\int \frac{x^{3}}{x^{3}-6 x^{2}+11 x-6} d x \\ &I=\int \frac{\left(x^{3}-6 x^{2}+11 x-6\right)-\left(-6 x^{2}+11 x-6\right)}{x^{3}-6 x+11 x-6} d x \end{aligned}$ [Add and subtract $-6x^{2}+11x+6$ ]
$\begin{aligned} &I=\int\left(1+\frac{6 x^{2}-11 x+6}{x^{3}-6 x+11 x-6}\right) d x \\ &I=\int d x+\int \frac{6 x^{2}-11 x+6}{(x-1)(x-2)(x-3)} d x \\ &I=x+I_{1} \end{aligned}$ (1)
Where
$\begin{aligned} &I_{1}=\int \frac{6 x^{2}-11 x+6}{(x-1)(x-2)(x-3)} d x \\ &\frac{6 x^{2}-11 x+6}{(x-1)(x-2)(x-3)}=\frac{A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)}{(x-1)(x-2)(x-3)} \\ &6 x^{2}+6-11 x=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2) \end{aligned}$
$\begin{aligned} &\text { At } x=2 \\ &6(4)+6-11(2)=A(0)+B(1)(-1)+C(0) \\ &30-22=-B \\ &8=-B \\ &B=-8 \end{aligned}$
$\begin{aligned} &\text { At } x=1 \\ &6(1)-11(1)+6=A(-1)(-2)+B(0)+C(0) \\ &1=2 A \\ &A=\frac{1}{2} \end{aligned}$
$\begin{aligned} &\text { At } x=3 \\ &6(9)-11(3)+6=A(0)+B(0)+C(2)(1) \\ &27=2 C \\ &C=\frac{27}{2} \end{aligned}$
$\begin{aligned} &\frac{6 x^{2}-11 x+6}{(x-1)(x-2)(x-3)}=\frac{1}{2(x-1)}-\frac{8}{x-2}+\frac{27}{2(x-3)} \\ &I_{1}=\frac{1}{2} \int \frac{1}{x-1} d x-8 \int \frac{1}{x-2} d x+\frac{27}{2} \int \frac{1}{x-3} d x \\ &I_{1}=\frac{1}{2} \log |x-1|-8 \log |x-2|+\frac{27}{2} \log |x-3|+C \end{aligned}$
Equation (1)
$I=x+\frac{1}{2} \log |x-1|-8 \log |x-2|+\frac{27}{2} \log |x-3|+C$

Indefinite Integrals exercise 18.30 question 11

Answer:
$\log \left|\frac{(\sin x+2)^{4}}{(\sin x+1)^{2}}\right|+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{\sin 2 x}{(1+\sin x)(2+\sin x)} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{\sin 2 x}{(1+\sin x)(2+\sin x)} d x \\ &I=\int \frac{2 \sin x \cos x}{(1+\sin x)(2+\sin x)} d x \quad[\sin 2 A=2 \sin A \cos A] \end{aligned}$
Let
$\begin{aligned} &\sin x=y \\ &\cos x d x=d y \\ &I=\int \frac{2 y d y}{(1+y)(2+y)} \end{aligned}$
$\begin{aligned} &\frac{2 y}{(1+y)(2+y)}=\frac{A}{1+y}+\frac{B}{2+y} \\ &\frac{2 y}{(1+y)(2+y)}=\frac{A(2+y)+B(1+y)}{(1+y)(2+y)} \\ &2 y=(2 A+B)+y(A+B) \end{aligned}$
Comparing coefficient
$2=A+B$ (1)
$2A+B=0$ (2)
Subtract equation (1) from equation (2)
$A=-2$
Equation (1)
$2=-2+B\\B=4$
Now
$\begin{aligned} &\frac{2 y}{(y+1)(2+y)}=\frac{-2}{y+1}+\frac{4}{y+2} \\ &I=-2 \int \frac{1}{y+1} d y+4 \int \frac{1}{y+2} d y \\ &I=-2 \log |y+1|+4 \log |y+2|+C \end{aligned}$
$\begin{aligned} &I=\log \left|\frac{(y+2)^{4}}{(y+1)^{2}}\right|+C \\ &I=\log \left|\frac{(\sin x+2)^{4}}{(\sin x+1)^{2}}\right|+C\quad\quad\quad\quad \quad[\because y=\sin x] \end{aligned}$

Indefinite Integrals exercise 18.30 question 12

Answer:
$\frac{1}{2} \log \left|\frac{x^{2}+1}{x^{2}+3}\right|+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{2 x}{\left(x^{2}+1\right)\left(x^{2}+3\right)} d x$
Explanation:
Let
$I=\int \frac{2 x}{\left(x^{2}+1\right)\left(x^{2}+3\right)} d x$
Let
$\begin{aligned} &x^{2}=y \\ &2 x d x=d y \\ &I=\int \frac{d y}{(y+1)(y+3)} \end{aligned}$
$\begin{aligned} &\frac{1}{(y+1)(y+3)}=\frac{A}{y+1}+\frac{B}{y+3} \\ &\frac{1}{(y+1)(y+3)}=\frac{A(y+3)+B(y+1)}{(y+1)(y+3)} \end{aligned}$
$1=A(y+3)+B(y+1)$ (1)
$\begin{aligned} &\text { At } y=-3 \text { equation }(1) \text { becomes }\\ &1=0+(-2) B\\ &B=\frac{1}{-2} \end{aligned}$
$\begin{aligned} &\text { At } y=-1 \text { equation (1) becomes }\\ &1=2 A+0\\ &A=\frac{1}{2} \end{aligned}$
$\begin{aligned} &\frac{1}{(y+1)(y+3)}=\frac{1}{2(y+1)}-\frac{1}{2(y+3)} \\ &I=\frac{1}{2} \int \frac{1}{y+1} d y-\frac{1}{2} \int \frac{1}{y+3} d y \end{aligned}$
$\begin{aligned} &I=\frac{1}{2} \log |y+1|-\frac{1}{2} \log |y+3|+C \\ &I=\frac{1}{2} \log \left|\frac{y+1}{y+3}\right|+C \\ &I=\frac{1}{2} \log \left|\frac{x^{2}+1}{x^{2}+3}\right|+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 13

Answer:
$\frac{1}{2} \log \left|\frac{\log x}{\log x+2}\right|+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{1}{x \log x(2+\log x)} d x$
Explanation:
$I=\int \frac{1}{x \log x(2+\log x)} d x$
Let
$\begin{aligned} &\log x=y \\ &\frac{1}{x}=d y \\ &I=\int \frac{d y}{y(y+2)} \end{aligned}$
Now
$\begin{aligned} &\frac{1}{y(y+2)}=\frac{A}{y}+\frac{B}{y+2} \\ &\frac{1}{y(y+2)}=\frac{A(y+2)+B y}{(y+2) y} \\ &1=A(2+y)+B y \\ &1=2 A+(A+B) y \end{aligned}$
Comparing the coefficient
$\begin{aligned} &2 A=1 \\ &A=\frac{1}{2} \\ &A+B=0 \\ &B=-A \\ &B=-A \\ &B=\frac{-1}{2} \end{aligned}$
$\begin{aligned} &\frac{1}{y(y+2)}=\frac{1}{2 y}-\frac{1}{2(y+2)} \\ &I=\frac{1}{2} \int \frac{1}{y} d y-\frac{1}{2} \int \frac{d y}{y+2} \end{aligned}$
$\begin{aligned} &I=\frac{1}{2} \log |y|-\frac{1}{2} \log |y+2|+C \\ &I=\frac{1}{2} \log \left|\frac{y}{y+2}\right|+C \\ &I=\frac{1}{2} \log \left|\frac{\log x}{\log x+2}\right|+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 14

Answer:
$\frac{3}{5} \log |x+2|+\frac{1}{5} \log \left|x^{2}+1\right|+\frac{1}{5} \tan ^{-1} x+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{x^{2}+x+1}{\left(x^{2}+1\right)(x+2)} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{x^{2}+x+1}{\left(x^{2}+1\right)(x+2)} d x \\ &\frac{x^{2}+x+1}{\left(x^{2}+1\right)(x+2)}=\frac{A}{x+2}+\frac{B x+C}{x^{2}+1} \end{aligned}$
$\begin{aligned} &\frac{x^{2}+x+1}{\left(x^{2}+1\right)(x+2)}=\frac{A\left(x^{2}+1\right)+(B x+C)(x+2)}{(x+2)\left(x^{2}+1\right)} \\ &x^{2}+x+1=A\left(x^{2}+1\right)+B x^{2}+2 B x+C x+2 C \\ &x^{2}+x+1=x^{2}(A+B)+(2 B+C) x+(A+2 C) \end{aligned}$
Comparing the coefficient
$A+B=1$ (1)
$2B+C=1$ (2)
$A+2C=1$ (3)
Subtract equation (3) from equation (1), we get
$\begin{aligned} &A+B=1 \quad- \\ &\frac{A+2 C=1}{B-2 C=0} \\ &B=2 C \end{aligned}$
Equation (2)
$\begin{aligned} &2(2 C)+C=1 \\ &4 C+C=1 \\ &5 C=1 \\ &C=\frac{1}{5} \\ &B=\frac{2}{5} \end{aligned}$
Equation (1)
$\begin{aligned} &A+\frac{2}{5}=1 \\ &A=1-\frac{2}{5} \\ &A=\frac{3}{5} \end{aligned}$
Now
$\frac{x^{2}+x+1}{\left(x^{2}+1\right)(x+2)}=\frac{3}{5(x+2)}+\frac{\frac{2}{5} x+\frac{1}{5}}{\left(x^{2}+1\right)}$
$\frac{x^{2}+x+1}{\left(x^{2}+1\right)(x+2)}=\frac{3}{5(x+2)}+\frac{2 x+1}{5\left(x^{2}+1\right)}$
$\begin{aligned} &I=\frac{3}{5} \int \frac{1}{x+2} d x+\frac{1}{5} \int \frac{2 x+1}{x^{2}+1} d x \\ &I=\frac{3}{5} \log |x+2|+\frac{1}{5} \int \frac{2 x}{x^{2}+1} d x+\frac{1}{5} \int \frac{1}{x^{2}+1} d x+C \\ &I=\frac{3}{5} \log |x+2|+\frac{1}{5} \log \left|x^{2}+1\right|+\frac{1}{5} \tan ^{-1} x+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 15

Answer:
$\! \! \! \! \! \! \! \! \! \! \! \frac{a\left(a^{2}+b\right)+c}{(a-b)(a-c)} \log |x-a|+\frac{b^{2}(a+1)+c}{(b-a)(b-c)} \log |x-b|+\frac{c(a c+b+1)}{(c-a)(c-b)} \log |x-c|+k$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{a x^{2}+b x+c}{(x-a)(x-b)(x-c)} d x \\$
Explanation:
$\begin{aligned} &I=\int \frac{a x^{2}+b x+c}{(x-a)(x-b)(x-c)} d x \\ &\frac{a x^{2}+b x+c}{(x-a)(x-b)(x-c)}=\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c} \end{aligned}$
$\begin{aligned} &\frac{a x^{2}+b x+c}{(x-a)(x-b)(x-c)}=\frac{A(x-b)(x-c)+B(x-a)(x-c)+C(x-a)(x-b)}{(x-a)(x-b)(x-c)} \\ &a x^{2}+b x+c=A(x-b)(x-c)+B(x-a)(x-c)+C(x-a)(x-b) \end{aligned}$
$\begin{aligned} &\text { For } x=b \\ &a b^{2}+b^{2}+c=0+B(b-a)(b-c)+(0) \\ &\frac{(a+1) b^{2}+c}{(b-a)(b-c)}=B \\ &\text { For } x=a \end{aligned}$

$\begin{aligned} &a^{3}+b a+c=A(a-b)(a-c)+0+0 \\ &\frac{a^{3}+a b+c}{(a-b)(a-c)}=A \end{aligned}$
$\begin{aligned} &\text { For } x=c \\ &a c^{2}+b c+c=0+0+C(c-a)(c-b) \\ &\frac{c(a c+b+1)}{(c-a)(c-b)}=C \end{aligned}$
$\begin{aligned} &\frac{a x^{2}+b x+c}{(x-a)(x-b)(x-c)}=\frac{a\left(a^{2}+b\right)+c}{(a-b)(a-c)} \cdot \frac{1}{x-a}+\frac{b^{2}(a+1)+c}{(b-a)(b-c)} \cdot \frac{1}{x-b}+\frac{c(a c+b+1)}{(c-a)(b-c)} \cdot \frac{1}{x-c} \\ &I=\frac{a\left(a^{2}+b\right)+c}{(a-b)(a-c)} \int \frac{1}{x-a} d x+\frac{b^{2}(a+1)+c}{(b-a)(b-c)} \int \frac{1}{x-b} d x+\frac{c(a c+b+1)}{(c-a)(c-b)} \int \frac{1}{x-c} d x \\ &I=\frac{a\left(a^{2}+b\right)+c}{(a-b)(a-c)} \log |x-a|+\frac{b^{2}(a+1)+c}{(b-a)(b-c)} \log |x-b|+\frac{c(a c+b+1)}{(c-a)(c-b)} \log |x-c|+k \end{aligned}$

Indefinite Integrals exercise 18.30 question 16

Answer:
$\frac{1}{2} \log |x-1|-\frac{1}{4} \log \left|x^{2}+1\right|+\frac{1}{2} \tan ^{-1} x+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{x}{\left(x^{2}+1\right)(x+1)} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{x}{\left(x^{2}+1\right)(x+1)} d x \\ &\frac{x}{\left(x^{2}+1\right)(x-1)}=\frac{A}{x-1}+\frac{B x+C}{x^{2}+1} \end{aligned}$
$\begin{aligned} &\frac{x}{\left(x^{2}+1\right)(x-1)}=\frac{A\left(x^{2}+1\right)+(B x+C)(x-1)}{(x-1)\left(x^{2}+1\right)} \\ &x=A\left(x^{2}+1\right)+B x^{2}-B x+C x-C \\ &x=(A+B) x^{2}+(C-B) x+A-C \end{aligned}$
Comparing the coefficient
$A+B=0$ (1)
$A=-B$
$C-B=1$ (2)
$A-C=0$ (3)
$A=C$
Equation (2)
$\begin{aligned} &A+A=1 \\ &2 A=1 \\ &A=\frac{1}{2} \\ &B=\frac{-1}{2} \\ &C=\frac{1}{2} \end{aligned}$
$\begin{aligned} &\frac{x}{\left(x^{2}+1\right)(x-1)}=\frac{1}{2(x-1)}+\frac{\frac{-1}{2} x+\frac{1}{2}}{\left(x^{2}+1\right)} \\ &=\frac{1}{2(x-1)}+\frac{1-x}{2\left(x^{2}+1\right)} \end{aligned}$
$\begin{aligned} &I=\frac{1}{2} \int \frac{1}{x-1} d x+\frac{1}{2} \int \frac{1-x}{x^{2}+1} d x \\ &I=\frac{1}{2} \int \frac{1}{x-1} d x+\frac{1}{2} \int \frac{1}{x^{2}+1} d x-\frac{1}{4} \int \frac{2 x}{x^{2}+1} d x \\ &I=\frac{1}{2} \log |x-1|+\frac{1}{2} \tan ^{-1} x-\frac{1}{4} \log \left|x^{2}+1\right|+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 17

Answer:
$\frac{1}{6} \log |x-1|-\frac{1}{2} \log |x+1|+\frac{1}{3} \log |x+2|+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{1}{(x-1)(x+1)(x+2)} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{1}{(x-1)(x+1)(x+2)} d x \\ &\frac{1}{(x-1)(x+2)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{x+2} \end{aligned}$
$\begin{aligned} &\frac{1}{(x-1)(x+2)(x+1)}=\frac{A(x+1)(x+2)+B(x-1)(x+2)+C(x-1)(x+1)}{(x-1)(x+2)(x+1)} \\ &1=A(x+1)(x+2)+B(x-1)(x+2)+C(x-1)(x+1) \end{aligned}$

$\begin{aligned} &\text { At } x=1 \\ &\begin{array}{l} 1=A(2)(3)+B(0)+C(0) \\ 1=6 A \\ A=\frac{1}{6} \end{array} \end{aligned}$
$\begin{aligned} &\text { At } x=-1 \\ &1=A(0)+B(-2)(1)+C(0) \\ &1=-2 B \\ &B=\frac{-1}{2} \end{aligned}$
$\begin{aligned} &\text { At } x=-2 \\ &1=A(0)+B(0)+C(-3)(-1) \\ &1=3 C \\ &C=\frac{1}{3} \end{aligned}$
$\begin{aligned} &\frac{1}{(x-1)(x+1)(x+2)}=\frac{1}{6(x-1)}-\frac{1}{2(x+1)}+\frac{1}{3(x+2)} \\ &I=\frac{1}{6} \int \frac{1}{x-1} d x-\frac{1}{2} \int \frac{1}{x+1} d x+\frac{1}{3} \int \frac{1}{x+2} d x \\ &I=\frac{1}{6} \log |x-1|-\frac{1}{2} \log |x+1|+\frac{1}{3} \log |x+2|+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 18

Answer:
$\frac{-2}{5} \tan ^{-1} \frac{x}{2}+\frac{3}{5} \tan \frac{x}{3}+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{x^{2}}{\left(x^{2}+4\right)\left(x^{2}+9\right)} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{x^{2}}{\left(x^{2}+4\right)\left(x^{2}+9\right)} d x \\ &x^{2}=(A+B x)\left(x^{2}+9\right)+(C x+D)\left(x^{2}+4\right) \\ &x^{2}=x^{3}(B+C)+x^{2}(A+D)+x(9 B+4 C)+(9 A+4 D) \end{aligned}$
Comparing the coefficient
$B+C=0$ (1)
$B=-C$
$A+D=1$ (2)
$9B+4C=0$ (3)
$9A+4D=1$ (4)
Equation (3)
$\begin{aligned} &9(-C)+4 C=0 \\ &-9 C+4 C=0 \\ &5 C=0 \\ &C=0 \\ &B=0 \end{aligned}$
Equation (2)
$A=1-D$
Equation (4)
$\begin{aligned} &9(1-D)+4 D=0 \\ &9-9 D+4 D=0 \\ &9=5 D \\ &D=\frac{9}{5} \\ &A=1-\frac{9}{5} \\ &A=\frac{-4}{5} \end{aligned}$
Now
$\begin{aligned} &\frac{x^{2}}{\left(x^{2}+4\right)\left(x^{2}+9\right)}=\frac{-4}{5\left(x^{2}+4\right)}+\frac{9}{5\left(x^{2}+9\right)} \\ &I=\frac{-4}{5} \int \frac{1}{x^{2}+4} d x+\frac{9}{5} \int \frac{1}{x^{2}+9} d x \end{aligned}$
$\begin{aligned} &I=\frac{-4}{5}\left(\frac{1}{2}\right) \tan ^{-1} \frac{x}{2}+\frac{9}{5}\left(\frac{1}{3}\right) \tan ^{-1} \frac{x}{3}+C \\ &I=\frac{-2}{5} \tan ^{-1} \frac{x}{2}+\frac{3}{5} \tan ^{-1} \frac{x}{3}+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 19

Answer:
$\log \left|\frac{\left(x^{2}-1\right)^{3}}{x}\right|+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{5 x^{2}-1}{x(x-1)(x+1)} d x$
Explanation:
Let
$I=\int \frac{5 x^{2}-1}{x(x-1)(x+1)} d x$
$\begin{aligned} &\frac{5 x^{2}+1}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1} \\ &5 x^{2}+1=A(x-1)(x+1)+B(x)(x+1)+C(x)(x-1) \\ &\text { At } x=1 \\ &5(1)+1=A(0)+B(1)(2)+C(0) \\ &6=2 B \\ &B=3 \end{aligned}$
$\begin{aligned} &\text { At } x=-1 \\ &5(1)+1=A(0)+B(0)+C(-1)(-2) \\ &6=2 C \\ &C=3 \end{aligned}$
$\begin{aligned} &\text { At } x=0 \\ &5(0)+1=A(-1)(1)+0+0 \\ &1=-A \\ &A=-1 \end{aligned}$
$\begin{aligned} &\frac{5 x^{2}-1}{x(x-1)(x+1)}=\frac{-1}{x}+\frac{3}{x-1}+\frac{3}{x+1} \\ &I=-\int \frac{1}{x} d x+3 \int \frac{1}{x-1} d x+3 \int \frac{1}{x+1} d x \\ &=-\log |x|+3 \log |x-1|+3 \log |x+1|+C \end{aligned}$
Using the formulas, $(x+y)(x-y)=x^{2}-y^{2} \& \log x+\log y=\log\left ( xy \right )$
$I=\log \left|\frac{\left(x^{2}-1\right)^{3}}{x}\right|+C$

Indefinite Integrals exercise 18.30 question 20

Answer:
$\log \left|\frac{x^{2}(x-2)}{(x+2)^{2}}\right|+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{x^{2}+6 x-8}{x^{3}-4 x} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{x^{2}+6 x-8}{x^{3}-4 x} d x \\ &I=\int \frac{x^{2}+6 x-8}{x\left(x^{2}-4\right)} d x \end{aligned}$
$\begin{aligned} &I=\int \frac{x^{2}+6 x-8}{x(x-2)(x+2)} d x \ldots .\left[x^{2}-y^{2}=(x+y)(x-y)\right] \\ &\frac{x^{2}+6 x-8}{x(x-2)(x+2)}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{x+2} \\ &x^{2}+6 x-8=A(x-2)(x+2)+B x(x+2)+C x(x-2) \end{aligned}$
$\begin{aligned} &\text { At } x=0 \\ &0+6(0)-8=A(-2)(2)+0+0 \\ &-8=-4 A \\ &A=2 \end{aligned}$
$\begin{aligned} &\text { At } x=2 \\ &4+6(2)-8=B(2)(4) \\ &8=8 B \\ &B=1 \end{aligned}$
$\begin{aligned} &\text { At } x=-2 \\ &4+6(-2)-8=A(0)+B(0)+C(-2)(-4) \\ &-16=8 C \\ &C=-2 \end{aligned}$
Now
$\begin{aligned} &\frac{x^{2}+6 x-8}{x(x-2)(x+2)}=\frac{2}{x}+\frac{1}{x-2}-\frac{2}{x+2} \\ &I=2 \int \frac{1}{x} d x+\int \frac{1}{x-2} d x-2 \int \frac{1}{x+2} d x \\ &I=2 \log |x|+\log |x-2|-2 \log |x+2|+C \\ &I=\log \left|\frac{x^{2}(x-2)}{(x+2)^{2}}\right|+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 21

Answer:
$\frac{1}{6} \log \left|\frac{(1+x)^{6}(-1+x)^{2}}{(2 x+1)^{5}}\right|+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{x^{2}+1}{(2 x+1)\left(x^{2}-1\right)} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{x^{2}+1}{(2 x+1)\left(x^{2}-1\right)} d x \\ &I=\int \frac{x^{2}+1}{(2 x+1)(x-1)(x+1)} d x \ldots\left[x^{2}-y^{2}=(x+y)(x-y)\right] \\ &\frac{x^{2}+1}{(2 x+1)(x-1)(x+1)}=\frac{A}{2 x+1}+\frac{B}{x-1}+\frac{C}{x+1} \\ &x^{2}+1=A(x-1)(x+1)+B(2 x+1)(x+1)+C(x-1)(2 x+1) \end{aligned}$
$\begin{aligned} &\text { At } x=1 \\ &1+1=A(0)+B(3)(2)+C(0) \\ &2=6 B \\ &B=\frac{1}{3} \end{aligned}$
$\begin{aligned} &B=\frac{1}{3} \\ &\text { At } x=-1 \\ &1+1=A(0)+B(0)+C(-1)(-2) \\ &2=2 C \\ &C=1 \end{aligned}$
$\begin{aligned} &\text { At } x=\frac{-1}{2} \\ &\frac{1}{4}+1=A\left(-\frac{1}{2}+1\right)\left(-\frac{1}{2}-1\right)+B(0)+C(0) \\ &\frac{5}{4}=A\left(\frac{-3}{4}\right) \\ &A=\frac{-5}{3} \end{aligned}$
$\begin{aligned} &\frac{x^{2}+1}{(2 x+1)(x-1)(x+1)}=\frac{-5}{3(2 x+1)}+\frac{1}{3(x-1)}+\frac{1}{(x+1)} \\ &I=\frac{-5}{3} \int \frac{1}{2 x-1} d x+\frac{1}{3} \int \frac{1}{x-1} d x+\int \frac{1}{x+1} d x \end{aligned}$
$\begin{aligned} &I=\frac{1}{3}\left[\frac{-5}{2} \log |2 x+1|+\log |x-1|+3 \log |x+1|\right]+C \\ &I=\frac{1}{6}[-5 \log |2 x+1|+2 \log |x-1|+6 \log |x+1|]+C \\ &I=\frac{1}{6} \log \left|\frac{(x-1)^{2}(x+1)^{6}}{(2 x+1)^{5}}\right|+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 22

Answer:
$\log \left|\frac{(2 \log x+1)^{\square}}{(3 \log x+2)^{1 / 2}}\right|+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{1}{x\left[6(\log x)^{2}+7 \log x+2\right]} d x$
Explanation:
$I=\int \frac{1}{x\left[6(\log x)^{2}+7 \log x+2\right]} d x$
Let
$\begin{aligned} &\log x=y \\ &\frac{1}{x} d x=d y \\ &I=\int \frac{d y}{6 y^{2}+7 y+2} d x \\ &I=\int \frac{d y}{(3 y+2)(2 y+1)} \end{aligned}$
$\begin{aligned} &\frac{1}{(3 y+2)(2 y+1)}=\frac{A}{3 y+2}+\frac{B}{2 y+1} \\ &1=A(2 y+1)+B(3 y+2) \\ &1=y(2 A+3 B)+(A+2 B) \end{aligned}$
Comparing the coefficient
$2A+3B=0$ (1)
$A+2B=1$ (2)
Multiply equation (2) by 2 and then
Subtract equation (1) from it
$\begin{aligned} &2 A+4 B=2 \quad- \\ &\frac{2 A+3 B=0}{B=2} \end{aligned}$
Equation (2)
$\begin{aligned} &A+2(2)=1 \\ &A+4=1 \\ &A=-3 \end{aligned}$
Now
$\begin{aligned} &\frac{1}{(3 y+2)(2 y+1)}=\frac{-3}{3 y+2}+\frac{2}{2 y+1} \\ &I=-3 \int \frac{1}{3 y+2} d y+2 \int \frac{1}{2 y+1} d y \\ &I=-3 \cdot \frac{1}{3} \log |3 y+2|+2 \cdot \frac{1}{2} \log |2 y+1| \\ &I=\log\left | \frac{2y+1}{3y+2} \right |+C \end{aligned}$
As $y=\log x$
$I=\log \left|\frac{(2 \log x+1)^{\square}}{(3 \log x+2)}\right|+C$

Indefinite Integrals exercise 18.30 question 23

Answer:
$\log \left|\frac{x^{n}}{x^{n}+1}\right|+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{1}{x\left(x^{n}+1\right)} d x$
Explanation:
Let
$\begin{aligned} &I=\int \frac{1}{x\left(x^{n}+1\right)} d x \\ &I=\int \frac{1}{x\left(x^{n}+1\right)} \cdot \frac{x^{n-1}}{x^{n-1}} d x \\ &I=\int \frac{x^{n-1}}{x^{n}\left(x^{n}+1\right)} d x \end{aligned}$
Let
$\begin{aligned} &x^{n}=y \\ &n x^{n-1} d x=d y \\ &x^{n-1} d x=\frac{d y}{n} \\ &I=\int \frac{d y}{n y(y+1)} \end{aligned}$
$\begin{aligned} &I=\frac{1}{n} \int \frac{d y}{y(y+1)} \\ &\frac{1}{y(y+1)}=\frac{A}{y}+\frac{B}{y+1} \\ &1=A(y+1)+B \end{aligned}$ (1)
At $y=0$ equation (1) becomes
$\begin{aligned} &1=A(1)+B(0) \\ &A=1 \end{aligned}$
At $y=-1$ equation (1) becomes

$\begin{aligned} &1=A(0)+B(-1) \\ &B=-1 \\ &\frac{1}{y(y+1)}=\frac{1}{y}-\frac{1}{y+1} \end{aligned}$
Thus
$\begin{aligned} &I=\int \frac{1}{y} d y-\int \frac{1}{y+1} d y \\ &I=\log |y|-\log |y+1|+C \end{aligned}$
As $y=x^{n}$
$\begin{aligned} &I=\log \left|x^{n}\right|-\log \left|x^{n}+1\right|+C \\ &I=\log \left|\frac{x^{n}}{x^{n}+1}\right|+C \mid \end{aligned}$

Indefinite Integrals exercise 18.30 question 24

Answer:
$\frac{1}{2\left(b^{2}-a^{2}\right)} \log \left|\frac{x-b}{x+a}\right|+\frac{1}{2\left(a^{2}-b^{2}\right)} \log \left|\frac{x-a}{x+b}\right|+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{x}{\left(x^{2}-a^{2}\right)\left(x^{2}-b^{2}\right)} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{x}{\left(x^{2}-a^{2}\right)\left(x^{2}-b^{2}\right)} d x \\ &I=\int \frac{x}{(x-a)(x-b)(x+a)(x+b)} d x \ldots\left(x^{2}-y^{2}\right)=(x+y)(x-y) \\ &\frac{x}{(x-a)(x+a)(x-b)(x+b)} d x=\frac{A}{x-a}+\frac{B}{x+a}+\frac{C}{x-b}+\frac{D}{x+b} \end{aligned}$
$\! \! \! \! \! \! \! \! \! \! x\! =\! A(x+a)(x-b)(x+b)\! +\! B(x-a)(x-b)(x+b)\! +\! C(x-a)(x+a)(x+b)\! +D\! (x-a)(x-b)(x+a)$ (1)
At $x=-a$ equation (1) becomes
$\begin{aligned} &-a=A(0)+B(-2 a)(-a-b)(-a+b)+C(0)+(0) \\ &-a=2 B a(a+b)(b-a) \\ &B=\frac{-1}{2\left(b^{2}-a^{2}\right)} \end{aligned}$
At $x=a$ equation (1) becomes
$\begin{aligned} &a=2 A(a)(a-b)(a+b)+B(0)+C(0)+D(0) \\ &1=2 A\left(a^{2}-b^{2}\right) \\ &A=\frac{1}{2\left(a^{2}-b^{2}\right)} \end{aligned}$
At $x=b$ equation (1) becomes
$\begin{aligned} &b=A(0)+B(0)+C(2 b)(b-a)(b+a) \\ &b=2 b C\left(b^{2}-a^{2}\right) \\ &C=\frac{1}{2\left(b^{2}-a^{2}\right)} \end{aligned}$
At $x=-b$ equation (1) becomes
$\begin{aligned} &-b=A(0)+B(0)+C(0)+(-2 b)(-b-a)(-b+a) \\ &-b=2 D b(b+a)(a-b) \\ &\, D=\frac{-1}{a^{2}-b^{2}} \end{aligned}$
$\begin{aligned} \frac{x}{(x-a)(x+a)(x-b)(x+b)} &=\frac{1}{2\left(a^{2}-b^{2}\right)(x-a)}+\frac{(-1)}{2\left(b^{2}-a^{2}\right)(x+a)}+\frac{1}{2\left(b^{2}-a^{2}\right)(x-b)}-\frac{1}{2\left(a^{2}-b^{2}\right)(x+b)} \\ \end{aligned}$
$\begin{aligned} &I=\frac{1}{2\left(a^{2}-b^{2}\right)} \int \frac{1}{x-a} d x+\frac{1}{2\left(b^{2}-a^{2}\right)} \int \frac{d x}{x-b}-\frac{1}{2\left(b^{2}-a^{2}\right)} \int \frac{d x}{x+a}-\frac{1}{2\left(a^{2}-b^{2}\right)} \int \frac{d x}{x+b} \\ &I=\frac{1}{2\left(a^{2}-b^{2}\right)} \log |x-a|+\frac{1}{2\left(b^{2}-a^{2}\right)} \log |x-b|-\frac{1}{2\left(b^{2}-a^{2}\right)} \log |x+a| \end{aligned}$
$\begin{gathered} I=\frac{1}{2\left(a^{2}-b^{2}\right)} \log |x-a|+\frac{1}{2\left(b^{2}-a^{2}\right)} \log |x-b|-\frac{1}{2\left(b^{2}-a^{2}\right)} \log |x+a| -\frac{1}{2\left(a^{2}-b^{2}\right)} \log |x+b|+C\\ \end{gathered}$
$I=\frac{1}{2\left(a^{2}-b^{2}\right)} \log \left|\frac{x-a}{x+b}\right|+\frac{1}{2\left(b^{2}-a^{2}\right)} \log \left|\frac{x-b}{x+a}\right|+C$

Indefinite Integrals exercise 18.30 question 25

Answer:
$\frac{-1}{14} \tan ^{-1} \frac{x}{2}+\frac{8}{35} \tan ^{-1} \frac{x}{5}+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{x^{2}+1}{\left(x^{2}+4\right)\left(x^{2}+25\right)} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{x^{2}+1}{\left(x^{2}+4\right)\left(x^{2}+25\right)} d x \\ &\frac{x^{2}+1}{\left(x^{2}+4\right)\left(x^{2}+25\right)}=\frac{A x+B}{x^{2}+4}+\frac{C x+D}{x^{2}+25} \\ &x^{2}+1=(A x+B)\left(x^{2}+25\right)+(C x+D)\left(x^{2}+4\right) \\ &x^{2}+1=x^{3}(A+C)+x^{2}(B+D)+x(25 A+4 C)+(25 B+4 D) \end{aligned}$
Comparing the coefficient
Coefficient of $x^{3}$
$A+C=0$ (2)
$A=-C$ (3)
Coefficient of $x^{2}$
$B+D=1$ (4)
Coefficient of $x$
$25A+4C=0$
$-21C+4C=0$ [From the equation (3)]
$-21C=0$
$C=0$ (5)
$A=0$ (6)
Constant term
$25 B+4 D=1$ (7)
Multiply the equation (4) by 4 and then subtract it from equation (7)
$\begin{aligned} &25 B+4 D=1- \\ &4 B+4 D=4 \\ &\overline{21 B=-3} \\ &B=\frac{-1}{7} \end{aligned}$
Equation (4)
$\begin{aligned} &\frac{-1}{7}+D=1 \\ &D=1+\frac{1}{7} \\ &D=\frac{8}{7} \\ &\frac{\left(x^{2}+1\right)}{\left(x^{2}+4\right)\left(x^{2}+25\right)}=\frac{-1}{7\left(x^{2}+4\right)}+\frac{8}{7\left(x^{2}+25\right)} \end{aligned}$
Thus
$\begin{aligned} &I=\frac{-1}{7} \int \frac{1}{x^{2}+4} d x+\frac{8}{7} \int \frac{1}{x^{2}+25} d x \\ &I=\frac{-1}{7} \cdot\left(\frac{1}{2}\right) \tan ^{-1} \frac{x}{2}+\frac{8}{7} \cdot\left(\frac{1}{5}\right) \tan ^{-1}\left(\frac{x}{5}\right)+C \\ &I=\frac{-1}{14} \tan ^{-1} \frac{x}{2}+\frac{8}{35} \tan ^{-1} \frac{x}{5}+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 26

Answer:
$\frac{x^{2}}{2}+\log \left|x^{2}-1\right|+\frac{1}{2} \log \left|\frac{x-1}{x+1}\right|+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{x^{3}+x+1}{x^{2}-1} d x$
Explanation:
Let
$\begin{aligned} &I=\int \frac{x^{3}+x+1}{x^{2}-1} d x \\ &I=\int\left(x+\frac{2 x+1}{x^{2}-1}\right) d x \\ &I=\int x d x+\int \frac{2 x}{x^{2}-1} d x+\int \frac{1}{x^{2}-1} d x \\ &I=\left[\frac{x^{2}}{2}\right]+\log \left|x^{2}-1\right|+\frac{1}{2} \log \left|\frac{x-1}{x+1}\right|+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 27

Answer:
$\frac{11}{4} \log \left|\frac{x+1}{x+3}\right|+\frac{5}{2(x+1)}+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{3 x-2}{(x+1)^{2}(x+3)} d x$
Explanation:
Let
$\begin{aligned} &I=\int \frac{3 x-2}{(x+1)^{2}(x+3)} d x \\ &\frac{3 x-2}{(x+1)^{2}(x+3)}=\frac{A}{(x+1)}+\frac{B}{(x+1)^{2}}+\frac{C}{x+3} \\ &\frac{3 x-2}{(x+1)^{2}(x+3)}=\frac{A(x+1)(x+3)+B(x+3)+C(x+1)^{2}}{(x+1)^{2}(x+3)} \end{aligned}$
$\begin{aligned} &3 x-2=A\left(x^{2}+4 x+3\right)+B(x+3)+C\left(x^{2}+1+2 x\right) \\ &3 x-2=(A+C) x^{2}+x(4 A+B+2 C)+(3 A+3 B+C) \end{aligned}$
Comparing the coefficient of $x^{2},x$ and constant term
$A+C=0$
$A=-C$ (1)
$4A+B+2C=3$ (2)
$-4C+B+2C=3$ from equation (1)
$B-2C=3$ (3)
$\begin{aligned} &3 A+3 B+C=-2 \\ &-3 C+3 B+C=-2 \end{aligned}$ from equation (1)
$3B-2C=-2$ (4)
Subtract equation (3) from equation (4)
$\begin{aligned} &3 B-2 C=-2- \\ &B-2 C=3 \\ &\overline{ 2 B=-5} \\ &B=\frac{-5}{2} \end{aligned}$
Equation (3)
$\begin{aligned} &\frac{-5}{2}-2 C=3 \\ &\frac{-5}{2}-3=2 C \\ &\frac{-11}{2}=2 C \\ &C=\frac{-11}{4} \\ &A=\frac{11}{4} \end{aligned}$
$\frac{3 x-2}{(x+1)^{2}(x+3)}=\frac{11}{4(x+1)}-\frac{5}{2(x+1)^{2}}-\frac{11}{4(x+3)}$
Thus
$\begin{aligned} &I=\frac{11}{4} \int \frac{1}{x+1} d x-\frac{5}{2} \int \frac{1}{(x+1)^{2}} d x-\frac{11}{4} \int \frac{1}{x+3} d x \\ &I=\frac{11}{4} \log |x+1|-\frac{5}{2}\left(\frac{1}{-1(x+1)}\right)-\frac{11}{4} \log |x+3|+C \\ &I=\frac{11}{4} \log \left|\frac{x+1}{x+3}\right|+\frac{5}{2(x+1)}+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 28

Answer:
$\frac{3}{25} \log \left|\frac{x-3}{x+2}\right|-\frac{7}{5(x-3)}+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{2 x+1}{(x+2)(x-3)^{2}} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{2 x+1}{(x+2)(x-3)^{2}} d x \\ &\frac{2 x+1}{(x+2)(x-3)^{2}}=\frac{A}{x+2}+\frac{B}{x-3}+\frac{C}{(x-3)^{2}} \\ &2 x+1=A(x-3)^{2}+B(x-3)(x+2)+C(x+2) \\ &2 x+1=x^{2}(A+B)+x(-6 A-B+C)+(9 A-6 B+2 C) \end{aligned}$
Comparing the coefficient of $x^{2},x$ and constant term
$A+B=0$ (1)
$\begin{aligned} &A=-B \\ &-6 A-B+C=2 \\ &-6 A+A+C=2 \end{aligned}$ [From equation 1]
$-5A+C=2$ (2)
$\begin{aligned} &9 A-6 B+2 C=1 \\ &9 A+6 A+2 C=1 \\ &15 A+2 C=1 \end{aligned}$ (3)
Multiply the equation (2) by 2 and subtract it from equation (3)
$\begin{aligned} &15 A+2 C=1\\ &\frac{-10 A+2 C=4}{25 A=-3}\\ &A=\frac{-3}{25}\\ &B=\frac{3}{25} \end{aligned}$ [From the equation (1)]
Equation (2)
$\begin{aligned} &-5\left(\frac{-3}{25}\right)+C=2 \\ &\frac{3}{5}+C=2 \\ &x=2-\frac{3}{5} \\ &C=\frac{7}{5} \end{aligned}$

$\begin{aligned} &\frac{2 x+1}{(x+2)(x-3)^{2}}=\frac{-3}{25(x+2)}+\frac{3}{25(x-3)}+\frac{7}{5(x-3)^{2}} \\ &I=\frac{-3}{25} \int \frac{1}{x+2} d x+\frac{3}{25} \int \frac{1}{x-3}+\frac{7}{5} \int \frac{1}{(x-3)^{2}} \\ &I=\frac{-3}{25} \log |x+2|+\frac{3}{25} \log |x-3|+\frac{7}{5} \cdot \frac{1}{(-1)(x-3)}+C \\ &I=\frac{3}{25} \log \left|\frac{x-3}{x+2}\right|-\frac{7}{5(x-3)}+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 29

Answer:
$\frac{3}{5} \log |x-2|+\frac{2}{5} \log |x+3|-\frac{1}{x-2}+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{x^{2}+1}{(x-2)^{2}(x+3)} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{x^{2}+1}{(x-2)^{2}(x+3)} d x \\ &\frac{x^{2}+1}{(x-2)^{2}(x+3)}=\frac{A}{x-2}+\frac{B}{(x-2)^{2}}+\frac{C}{x+3} \end{aligned}$
$\begin{aligned} &x^{2}+1=A(x-2)(x+3)+B(x+3)+C(x-2)^{2} \\ &x^{2}+1=A\left(x^{2}+x-6\right)+B(x+3)+C\left(x^{2}+4-4 x\right) \\ &x^{2}+1=(A+C) x^{2}+(A+B-4 C) x+(-6 A+3 B+4 C) \end{aligned}$
Equating the similar terms, we get
$A+C=1$ (1)
$A+B-4C=0$ (2)
$-6A+3B+4C=1$ (3)
Subtract equation (1) from equation (2) and we get
$B-5C=-1$ (4)
Multiply equation (1) by 6 and then adding equation (3)
$3B+10C=7$ (5)
Multiply equation (4) by 3 and then subtract it from equation (5)
$\! \! \! \! \! \! \! \! 3 B+10 C=7 \\ 3 B-15 C=-3 \\ \overline{25 C=10 }\\\\$
$\begin{aligned} &C=\frac{10}{25} \\ &C=\frac{2}{5} \end{aligned}$
Equation (4)
$\begin{aligned} &B-5\left(\frac{2}{5}\right)=-1 \\ &B-2=-1 \\ &B=1 \end{aligned}$
Equation (1)
$\begin{aligned} &A+\frac{2}{5}=1 \\ &A=1-\frac{2}{5} \\ &A=\frac{3}{5} \end{aligned}$
Now
$\frac{x^{2}+1}{(x-2)^{2}(x+3)}=\frac{3}{5(x-2)}+\frac{1}{(x-2)^{2}}+\frac{2}{5(x+3)}$
Thus
$\begin{aligned} &I=\frac{3}{5} \int \frac{1}{x-2} d x+\int \frac{1}{(x-2)^{2}} d x+\frac{2}{5} \int \frac{1}{x+3} d x \\ &I=\frac{3}{5} \log |x-2|-\frac{1}{x-2}+\frac{2}{5} \log |x+3|+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 30

Answer:
$\frac{2}{9} \log \left|\frac{x-1}{x+2}\right|-\frac{1}{3(x-1)}+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{x}{(x-1)^{2}(x+2)} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{x}{(x-1)^{2}(x+2)} d x \\ &\frac{x}{(x-1)^{2}(x+2)}=\frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{C}{x+2} \\ &x=A(x-1)(x+2)+B(x+2)+C(x-1)^{2} \\ &x=x^{2}(A+C)+x(A+B-2 C)+(-2 A+2 B+C) \end{aligned}$
Equating the similar terms
$\begin{aligned} &A+C=0 \\ &A=-C \end{aligned}$ (1)
$A+B-2C=1$ (2)
$\begin{aligned} &A+B+2 A=1 \\ &B+3 A=1 \end{aligned}$ (3)
$\begin{aligned} &-2 A+2 B+C=0 \\ &-2 A+2 B-A=0 \\ &2 B-3 A=0 \end{aligned}$ (4)
Adding equation (3) and (4)
$\begin{aligned} &3 B=1 \\ &B=\frac{1}{3} \end{aligned}$
Equation (3)
$\begin{aligned} &\frac{1}{3}+3 A=1 \\ &3 A=1-\frac{1}{3} \\ &3 A=\frac{2}{3} \\ &A=\frac{2}{9} \\ &C=\frac{-2}{9} \end{aligned}$
$\frac{x}{(x-1)^{2}(x+2)}=\frac{2}{9(x-1)}+\frac{1}{3(x-1)^{2}}+\frac{-2}{9(x+2)}$
$\begin{aligned} &I=\frac{2}{9} \int \frac{d x}{x-1}-\frac{2}{9} \int \frac{d x}{x+2}+\frac{1}{3} \int \frac{d x}{(x-1)^{2}} \\ &I=\frac{2}{9} \log |x-1|-\frac{2}{9} \log |x+2|-\frac{1}{3(x-1)}+C \\ &I=\frac{2}{9} \log \left|\frac{x-1}{x+2}\right|-\frac{1}{3(x-1)}+C \end{aligned}$

Indefinite Integrals Excercise 18.30 Question 31

Answer:
$\frac{1}{4} \log |x-1|+\frac{3}{4} \log |x+1|+\frac{1}{2(x+1)}+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{x^{2}}{(x-1)(x+1)^{2}} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{x^{2}}{(x-1)(x+1)^{2}} d x \\ &\frac{x^{2}}{(x-1)(x+1)^{2}}=\frac{A}{x+1}+\frac{B}{x-1}+\frac{C}{(x+1)^{2}} \end{aligned}$
$\begin{aligned} &x^{2}=A(x+1)(x-1)+B(x+1)^{2}+C(x-1) \\ &x^{2}=x^{2}(A+B)+x(2 B+C)+(-A+B-C) \end{aligned}$
Equating similar terms
$A+B=1$ (1)
$2B+C=0$ (2)
$\begin{aligned} &C=-2 B \\ &-A+B-C=0 \end{aligned}$ (3)
$\begin{aligned} &-A+B+2 B=0 \\ &-A+3 B=0 \end{aligned}$ (4)
Adding equation (1) and (4)
$\begin{gathered} 4 B=1 \\ B=\frac{1}{4} \end{gathered}$
Equation (2)
$\begin{aligned} &C=-2 \times \frac{1}{4} \\ &C=\frac{-1}{2} \end{aligned}$
Equation (1)
$\begin{aligned} &A=1-\frac{1}{4} \\ &A=\frac{3}{4} \end{aligned}$
$\begin{aligned} &\frac{x^{2}}{(x-1)(x+1)^{2}}=\frac{3}{4(x+1)}+\frac{1}{4(x-1)}-\frac{1}{2(x+1)^{2}} \\ &I=\frac{3}{4} \int \frac{1}{x+1} d x+\frac{1}{4} \int \frac{1}{x-1} d x-\frac{1}{2} \int \frac{1}{(x+1)^{2}} d x \\ &I=\frac{3}{4} \log |x+1|+\frac{1}{4} \log |x-1|+\frac{1}{2(x+1)}+C \end{aligned}$

Indefinite Integrals Excercise 18.30 Question 32

Answer:
$\frac{1}{x+1}+\log |x+2|+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{x^{2}+x-1}{(x+1)^{2}(x+2)} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{x^{2}+x-1}{(x+1)^{2}(x+2)} d x \\ &\frac{x^{2}+x-1}{(x+1)^{2}(x+2)}=\frac{A}{x+1}+\frac{B}{(x+1)^{2}}+\frac{C}{x+2} \\ &x^{2}+x-1=A(x+1)(x+2)+B(x+2)+C(x+1)^{2} \\ &x^{2}+x-1=(A+C) x^{2}+(3 A+B+2 C) x+(2 A+2 B+C) \end{aligned}$
Equating the similar terms
$\begin{aligned} &A+C=1 \\ &3 A+B+2 C=1 \\ &2 A+2 B+C=-1 \end{aligned}$
On solving we get
$A=0 ,B=-1, C=1$
Thus
$\begin{aligned} &\frac{x^{2}+x-1}{(x+1)^{2}(x+2)}=\frac{0}{x+1}+\frac{(-1)}{(x+1)^{2}}+\frac{1}{x+2} \\ &I=\int \frac{-1}{(x+1)^{2}} x+\int \frac{1}{x+2} d x \\ &I=\frac{1}{x+1}+\log |x+2|+C \end{aligned}$

Indefinite Integrals Excercise 18.30 Question 33

Answer:
$13 \log |x|+\frac{13}{x}-12 \log |2 x+1|+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{2 x^{2}+7 x-3}{x^{2}(2 x+1)} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{2 x^{2}+7 x-3}{x^{2}(2 x+1)} d x \\ &\frac{2 x^{2}+7 x-3}{x^{2}(2 x+1)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{2 x+1} \\ &2 x^{2}+7 x-3=A x(2 x+1)+B(2 x+1)+C x^{2} \\ &2 x^{2}+7 x-3=x^{2}(2 A+C)+x(A+2 B)+B \end{aligned}$
Equating similar terms
$2A+C=2$ (1)
$A+2B=7$ (2)
$B=-3$ (3)
Equation (2)
$\begin{aligned} &A-6=7 \\ &A=13 \end{aligned}$
Equation (1)
$\begin{aligned} &26+C=2 \\ &C=-24 \\ &\frac{2 x^{2}+7 x-3}{x^{2}(x+1)}=\frac{13}{x}-\frac{3}{x^{2}}-\frac{24}{2 x+1} \end{aligned}$
$\begin{aligned} &I=13 \int \frac{d x}{x}-13 \int \frac{d x}{x^{2}}-24 \int \frac{d x}{2 x+1} \\ &I=13 \log |x|+\frac{13}{x}-24 \cdot\left(\frac{1}{2}\right) \log |2 x+1|+C \\ &I=13 \log |x|+\frac{13}{x}-12 \log |2 x+1|+C \end{aligned}$

Indefinite Integrals Excercise 18.30 Question 34

Answer:
$6 \log |x|-\log |x+1|-\frac{9}{x+1}+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{5 x^{2}+20 x+6}{x^{3}+2 x^{2}+x} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{5 x^{2}+20 x+6}{x^{3}+2 x^{2}+x} d x \\ &I=\int \frac{5 x^{2}+20 x+6}{x(x+1)^{2}} d x \end{aligned}$
$\begin{aligned} &\frac{5 x^{2}+20 x+6}{x(x+1)^{2}}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^{2}} \\ &5 x^{2}+20 x+6=A(x+1)^{2}+B(x)(x+1)+C(x) \\ &5 x^{2}+20 x+6=x^{2}(A+B)+x(2 A+B+C)+A \end{aligned}$
Equating the similar terms
$5=A+B$ (1)
$A=6$ (2)
Equation (1)
$\begin{aligned} &5=6+B \\ &B=-1 \\ &2 A+B+C=20 \\ &12-1+C=20 \\ &11+C=20 \\ &C=9 \end{aligned}$
$\begin{aligned} &\frac{5 x^{2}+20 x+6}{x(x+1)^{2}}=\frac{6}{x}-\frac{1}{x+1}+\frac{9}{(x+1)^{2}} \\ &I=6 \int \frac{d x}{x}-\int \frac{d x}{x+1}+9 \int \frac{d x}{(x+1)^{2}} \\ &I=6 \log |x|-\log |x+1|-\frac{9}{x+1}+C \end{aligned}$

Indefinite Integrals Excercise 18.30 Question 35

Answer:
$\frac{9}{4} \log |x+2|-\frac{9}{8} \log \left|x^{2}+4\right|+\frac{9}{4} \tan ^{-1} \frac{x}{2}+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{18}{(x+2)\left(x^{2}+4\right)} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{18}{(x+2)\left(x^{2}+4\right)} d x \\ &\frac{18}{(x+2)\left(x^{2}+4\right)}=\frac{A}{x+2}+\frac{B x+C}{x^{2}+4} \\ &18=A\left(x^{2}+4\right)+(B x+C)(x+2) \\ &18=x^{2}(A+B)+(2 B+C) x+(4 A+2 C) \end{aligned}$
Equating the similar terms
$\begin{aligned} &A+B=0 \\ &A=-B \end{aligned}$ (1)
$\begin{aligned} &2 B+C=0 \\ &C=-2 B \end{aligned}$ (2)
$\begin{aligned} &4 A+2 C=18 \\ &-4 B-4 B=18 \\ &-8 B=18 \\ &B=\frac{-9}{4} \end{aligned}$
Equation (2)
$\begin{aligned} &C=-2 \times\left(\frac{-9}{4}\right) \\ &C=\frac{9}{2} \end{aligned}$
Equation (1)
$\begin{aligned} A=\frac{9}{4} \\ \end{aligned}$
$\begin{aligned} &\frac{18}{(x+2)\left(x^{2}+4\right)}=\frac{9}{4(x+2)}+\frac{\frac{-9}{4} x+\frac{9}{2}}{x^{2}+4} \\ &=\frac{9}{4(x+2)}+\frac{-9 x+18}{4\left(x^{2}+4\right)} \\ &I=\frac{9}{4} \int \frac{d x}{x+2}-\frac{9}{8} \int \frac{2 x}{x^{2}+4} d x+\frac{9}{2} \int \frac{1}{x^{2}+4} d x \end{aligned}$
$\frac{9}{4} \log |x+2|-\frac{9}{8} \log \left|x^{2}+4\right|+\frac{9}{4} \tan ^{-1} \frac{x}{2}+C$

Indefinite Integrals exercise 18.30 question 36

Answer:
$\frac{-1}{2} \log \left|x^{2}+1\right|+2 \tan ^{-1} x+\log |x+2|+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{5}{\left(x^{2}+1\right)(x+2)} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{5}{\left(x^{2}+1\right)(x+2)} d x \\ &\frac{5}{\left(x^{2}+1\right)(x+2)}=\frac{A x+B}{x^{2}+1}+\frac{C}{x+2} \\ &5=(A x+B)(x+2)+C\left(x^{2}+1\right) \\ &5=x^{2}(A+C)+x(2 A+B)+(2 B+C) \end{aligned}$
Equating the similar terms
$\begin{aligned} &A+C=0 \\ &A=-C \end{aligned}$ (1)
$\begin{aligned} &2 A+B=0 \\ &B=-2 A \\ &2 B+C=5 \\ &-4 A-A=5 \end{aligned}$ [From equation (1) and equation (2)]
$\begin{aligned} &-5 A=5 \\ &A=-1 \end{aligned}$
Equation (2)
$B=2$
Equation (1)
$\begin{aligned} &C=1 \\ &\frac{5}{\left(x^{2}+1\right)(x+2)}=\frac{-x+2}{x^{2}+1}+\frac{1}{x+2} \\ &I=\int \frac{2-x}{x^{2}+1} d x+\int \frac{1}{x+2} d x \\ &I=2 \int \frac{1}{x^{2}+1} d x-\frac{1}{2} \int \frac{2 x d x}{x^{2}+1}+\int \frac{1}{x+2} d x \\ &I=2 \tan ^{-1} x-\frac{1}{2} \log \left|x^{2}+1\right|+\log |x+2|+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 37

Answer:
$\frac{-1}{2} \log |x+1|+\frac{1}{4} \log \left|x^{2}+1\right|+\frac{1}{2} \tan ^{-1} x+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{x}{(x+1)\left(x^{2}+1\right)} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{x}{(x+1)\left(x^{2}+1\right)} d x \\ &\frac{x}{(x+1)\left(x^{2}+1\right)}=\frac{A}{x+1}+\frac{B x+C}{x^{2}+1} \\ &x=A\left(x^{2}+1\right)+(B x+C)(x+1) \\ &x=x^{2}(A+B)+(B+C) x+(A+C) \end{aligned}$

Equating similar terms

$\begin{aligned} &A+B=0 \\ &B=-A \end{aligned}$ (1)
$\begin{aligned} &A+C=0 \\ &C=-A \end{aligned}$ (2)
$\begin{aligned} &B+C=1 \\ &-A-A=1 \\ &-2 A=1 \\ &A=\frac{-1}{2} \end{aligned}$
Equation (1)
$B=\frac{1}{2}$
Equation (2)
$\begin{aligned} &C=\frac{1}{2} \\ &\frac{x}{(x+1)\left(x^{2}+1\right)}=\frac{-1}{2(x+1)}+\frac{\frac{1}{2} x+\frac{1}{2}}{x^{2}+1} \\ &=\frac{-1}{2(x+1)}+\frac{x+1}{2\left(x^{2}+1\right)} \end{aligned}$
$\begin{aligned} &I=\frac{-1}{2} \int \frac{1}{x+1} d x+\frac{1}{2} \int \frac{x}{x^{2}+1} d x+\frac{1}{2} \int \frac{1}{x^{2}+1} d x \\ &I=\frac{-1}{2} \log |x+1|+\frac{1}{4} \log \left|x^{2}+1\right|+\frac{1}{2} \tan ^{-1} x+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 38

Answer:
$\frac{-1}{4} \log \left|x^{2}+1\right|+\frac{1}{2} \tan ^{-1} x+\frac{1}{2} \log |x+1|+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{1}{1+x+x^{2}+x^{3}} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{1}{1+x+x^{2}+x^{3}} d x \\ &I=\int \frac{1}{(1+x)+x^{2}(x+1)} d x \\ &I=\int \frac{d x}{(1+x)\left(1+x^{2}\right)} \end{aligned}$
$\begin{aligned} &\frac{1}{\left(1+x^{2}\right)(1+x)}=\frac{A x+B}{1+x^{2}}+\frac{C}{1+x} \\ &1=(A x+B)(1+x)+C\left(1+x^{2}\right) \\ &1=x^{2}(A+C)+x(B+A)+(B+C) \end{aligned}$
Equating similar terms
$\begin{aligned} &A+C=0 \\ &A=-C \end{aligned}$ (1)
$\begin{aligned} &B+A=0 \\ &A=-B \end{aligned}$ (2)
$\begin{aligned} &B+C=1 \\ &-A-A=1 \\ &-2 A=1 \\ &A=\frac{-1}{2} \end{aligned}$
Equation (1)
$C=\frac{1}{2}$
Equation (2)
$B=\frac{1}{2}$
$\begin{aligned} &\frac{1}{1+x+x^{2}+x^{3}}=\frac{-\frac{1}{2} x+\frac{1}{2}}{x^{2}+1}+\frac{\frac{1}{2}}{x+1} \\ &=\frac{-x+1}{2\left(x^{2}+1\right)}+\frac{1}{2(x+1)} \end{aligned}$
$\begin{aligned} &I=-\frac{1}{2} \int \frac{x}{x^{2}+1} d x+\frac{1}{2} \int \frac{1}{x^{2}+1} d x+\frac{1}{2} \int \frac{1}{x+1} d x \\ &I=\frac{-1}{4} \log \left|x^{2}+1\right|+\frac{1}{2} \tan ^{-1} x+\frac{1}{2} \log |x+1|+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 39

Answer:
$\frac{1}{2} \log |x+1|-\frac{1}{2(x+1)}-\frac{1}{4} \log \left|x^{2}+1\right|+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{1}{(x+1)^{2}\left(x^{2}+1\right)} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{1}{(x+1)^{2}\left(x^{2}+1\right)} d x \\ &\frac{1}{(x+1)^{2}\left(x^{2}+1\right)}=\frac{A}{(x+1)}+\frac{B}{(x+1)^{2}}+\frac{C x+D}{x^{2}+1} \end{aligned}$
$\begin{aligned} &1=A(x+1)\left(x^{2}+1\right)+B\left(x^{2}+1\right)+(C x+D)(x+1)^{2} \\ &1=(A+C) x^{3}+(A+B+2 C+D) x^{2}+(A+C+2 D) x+(A+B+D) \end{aligned}$
Equating the similar terms
$\begin{aligned} &A+C=0 \\ &C=-A \end{aligned}$ (1)
$\begin{aligned} &A+B+2 C+D=0 \\ &A+B-2 A+D=0 \end{aligned}$ [From equation (1)]
$-A+B+D=0$ (2)
$A+C+2D=0$
$2D=0$ [From equation (1)]
$D=0$ (3)
Equation (2)
$\begin{aligned} &-A+B=0 \\ &A=B \end{aligned}$ (4)
$\begin{aligned} &A+B+D=1 \\ &A+A+0=1 \end{aligned}$ [From equation (4) and (3)]
$\begin{aligned} &2 A=1 \\ &A=\frac{1}{2} \end{aligned}$
Equation (4)
$B=\frac{1}{2}$
And equation (1)
$C=\frac{-1}{2}$
$\begin{aligned} &\frac{1}{(x+1)^{2}\left(x^{2}+1\right)}=\frac{1}{2(x+1)}+\frac{1}{2(x+1)^{2}}-\frac{1 x}{2\left(x^{2}+1\right)} \\ &I=\frac{1}{2} \int \frac{1}{x+1} d x+\frac{1}{2} \int \frac{1}{(x+1)^{2}}-\frac{1}{2} \int \frac{x}{\left(x^{2}+1\right)} d x \\ &I=\frac{1}{2} \log |x+1|-\frac{1}{2(x+1)}-\frac{1}{4} \log \left|x^{2}+1\right|+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 40

Answer:
$\frac{2}{3} \log |x-1|-\frac{1}{3} \log \left|x^{2}+x+1\right|+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{2 x}{x^{3}-1} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{2 x}{x^{3}-1} d x \\ &I=\int \frac{2 x}{(x-1)\left(x^{2}+x+1\right)} d x \quad \quad \quad \quad \quad \quad \quad\left[a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)\right] \end{aligned}$
$\begin{aligned} &\frac{2 x}{(x+1)\left(x^{2}+x+1\right)}=\frac{A}{x-1}+\frac{B x+C}{x^{2}+x+1} \\ &2 x=A\left(x^{2}+x+1\right)+(B x+C)(x-1) \\ &2 x=x^{2}(A+B)+x(A-B+C)+(A-C) \end{aligned}$
Equating the similar terms
$\begin{aligned} &A+B=0 \\ &A=-B \end{aligned}$ (1)
$\begin{aligned} &A-C=0 \\ &A=C \end{aligned}$ (2)
$\begin{aligned} & A-B+C=2 \\ &A+A+A=2 \end{aligned}$ [From equation (1) and (2)]
$\begin{aligned} &3 A=2 \\ &A=\frac{2}{3} \end{aligned}$
Equation (1)
$B=\frac{-2}{3}$
Equation (2)
$C=\frac{2}{3}$
$\begin{aligned} &\frac{2 x}{(x-1)\left(x^{2}+x+1\right)}=\frac{2}{3(x-1)}+\frac{\frac{-2}{3} x+\frac{2}{3}}{x^{2}+x+1} \\ &=\frac{2}{3(x-1)}+\frac{2-2 x}{3\left(x^{2}+2 x+1\right)} \end{aligned}$
$\begin{aligned} &I=\frac{2}{3} \int \frac{1}{x-1} d x-\frac{1}{3} \int \frac{2 x-2}{\left(x^{2}+x+1\right)} d x \\ &I=\frac{2}{3} \int \frac{1}{x-1} d x-\frac{1}{3} \int \frac{2 x+1}{x^{2}+x+1} d x-\frac{1}{3} \int \frac{-3 d x}{x^{2}+x+1} d x \\ &I=\frac{2}{3} \int \frac{1}{x-1} d x-\frac{1}{3} \int \frac{2 x+1}{x^{2}+x+1} d x+\int \frac{d x}{\left(x+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} d x \end{aligned}$
$I=\frac{2}{3} \log |x-1|-\frac{1}{3} \log \left|x^{2}+x+1\right|+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+C$
$I=\frac{2}{3} \log |x-1|-\frac{1}{3} \log \left|x^{2}+x+1\right|+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+C$

Indefinite Integrals exercise 18.30 question 41

Answer:
$\frac{1}{3} \tan ^{-1} x-\frac{1}{6} \tan ^{-1} \frac{x}{2}+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)} d x \\ &\frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac{A x+B}{x^{2}+1}+\frac{C x+D}{x^{2}+4} \\ &1=(A x+B)\left(x^{2}+4\right)+(C x+D)\left(x^{2}+1\right) \\ &1=x^{3}(A+C)+x^{2}(B+D)+(4 A+C) x+4 B+D \end{aligned}$
Equating the similar terms
$\begin{aligned} &A+C=0 \\ &A=-C \end{aligned}$ (1)
$\begin{aligned} &B+D=0 \\ &B=-D \end{aligned}$ (2)
$\begin{aligned} &4 A+C=0 \\ &-4 C+C=0 \end{aligned}$ [From the equation (1)]
$\begin{aligned} &-3 C=0 \\ &C=0 \\ &A=0 \end{aligned}$ [From the equation (1)]
$\begin{aligned} &4 B+D=1 \\ &-4 D+D=1 \\ &-3 D=1 \end{aligned}$
$D=\frac{-1}{3}$ (3)
$B=\frac{1}{3}$ [From the equation (2)]
$\begin{aligned} &\frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac{1}{3\left(x^{2}+1\right)}+\frac{(-1)}{3\left(x^{2}+4\right)} \\ &I=\frac{1}{3} \int \frac{1}{x^{2}+1} d x-\frac{1}{3} \int \frac{1}{x^{2}+4} d x \\ &I=\frac{1}{3} \tan ^{-1} x-\frac{1}{3} \cdot\left(\frac{1}{2}\right) \tan ^{-1} \frac{x}{2}+C \\ &I=\frac{1}{3} \tan ^{-1} x-\frac{1}{6} \tan ^{-1} \frac{x}{2}+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 42

Answer:
$\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{\sqrt{3} x}{2}\right)-\tan ^{-1} x+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{x^{2}}{\left(x^{2}+1\right)\left(3 x^{2}+4\right)} d x \\$
Explanation:
Let

$\begin{aligned} &I=\int \frac{x^{2}}{\left(x^{2}+1\right)\left(3 x^{2}+4\right)} d x \\ &\frac{x^{2}}{\left(x^{2}+1\right)\left(3 x^{2}+4\right)}=\frac{A x+B}{x^{2}+1}+\frac{C x+D}{3 x^{2}+4} \\ &x^{2}=(A x+B)\left(3 x^{2}+4\right)+(C x+D)\left(x^{2}+1\right) \\ &x^{2}=(3 A+C) x^{3}+(3 B+D) x^{2}+(4 A+C) x+(4 B+D) \end{aligned}$
Equating the similar term
$\begin{aligned} &3 A+C=0 \\ &3 A=-C \end{aligned}$ (1)
$4A+C= 0$
$4A-3A= 0$ [From equation (1)]
$A= 0$ (2)
Equation (1)
$\begin{aligned} &C=0 \\ &3 B+D=1 \end{aligned}$ (3)
$\begin{aligned} &4 B+D=0 \\ &D=-4 B \end{aligned}$ (4)
Equation (3)
$\begin{aligned} &3 B-4 B=1 \\ &-B=1 \\ &B=-1 \\ &D=4 \end{aligned}$ [From equation (4)]
$\begin{aligned} &I=-1 \int \frac{1}{x^{2}+1} d x+4 \int \frac{1}{3 x^{2}+4} d x \\ &I=-\tan ^{-1} x+\frac{4}{3} \int \frac{1}{x^{2}+\frac{4}{3}} d x+C \end{aligned}$
$\begin{aligned} &I=-\tan ^{-1} x+\frac{4}{3} \int \frac{1}{x^{2}+\left(\frac{2}{\sqrt{3}}\right)^{2}} d x+C \\ &I=-\tan ^{-1} x+\frac{4}{3}\left(\frac{\sqrt{3}}{2}\right) \tan ^{-1}\left(\frac{x}{\frac{2}{\sqrt{3}}}\right)+C \\ &I=-\tan ^{-1} x+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{\sqrt{3} x}{2}\right)+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 43

Answer:
$\frac{1}{2} \log \left|\frac{x+1}{x-1}\right|-\frac{4}{x-1}+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{3 x+5}{x^{3}-x^{2}-x+1} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{3 x+5}{x^{3}-x^{2}-x+1} d x \\ &I=\int \frac{3 x+5}{x^{2}(x-1)-1(x-1)} d x \\ &I=\int \frac{3 x+5}{\left(x^{2}-1\right)(x-1)} d x \end{aligned}$
$\begin{aligned} &I=\int \frac{3 x+5}{(x+1)(x-1)^{2}} d x \\ &\frac{3 x+5}{(x+1)(x-1)^{2}}=\frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{C}{x+1} \\ &3 x+5=A(x-1)(x+1)+B(x+1)+C(x-1)^{2} \end{aligned}$ (1)
Put $x= 1$ in equation (1)
$\begin{aligned} &3+5=A(0)+B(2)+C(0) \\ &8=2 B \\ &B=4 \end{aligned}$
Put $x= -1$ in equation (1)
$\begin{aligned} &-3+5=A(0)+B(0)+C(4) \\ &2=4 C \\ &C=\frac{1}{2} \end{aligned}$
Put $x=0$ in equation (1)
$\begin{aligned} &5=A(-1)(1)+B+C \\ &5=-A+B+C \\ &5=-A+4+\frac{1}{2} \\ &1=-A+\frac{1}{2} \end{aligned}$
$\begin{aligned} &\frac{1}{2}=-A \\ &A=\frac{-1}{2} \\ &\frac{3 x+5}{(x+1)(x-1)^{2}}=\frac{-1}{2(x-1)}+\frac{4}{(x-1)^{2}}+\frac{1}{2(x+1)} \end{aligned}$
Thus
$\begin{aligned} &I=\frac{-1}{2} \int \frac{d x}{x-1}+4 \int \frac{d x}{(x-1)^{2}}+\frac{1}{2} \int \frac{d x}{x+1} \\ &I=\frac{-1}{2} \log |x-1|+(-4) \frac{1}{x-1}+\frac{1}{2} \log |x+1|+C \\ &I=\frac{1}{2} \log \left|\frac{x+1}{x-1}\right|-\frac{4}{x-1}+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 44

Answer:
$x-\log |x|+\frac{1}{2} \log \left|x^{2}+1\right|-\tan ^{-1} x+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{x^{3}-1}{x^{3}+x} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{x^{3}-1}{x^{3}+x} d x \\ &I=\int \frac{x^{3}+x-x-1}{x^{3}+x} d x \\ &I=\int\left[\frac{x^{3}+x}{x^{3}+x}-\frac{x+1}{x^{3}+x}\right] d x \end{aligned}$
$\begin{aligned} &I=\int\left[1-\frac{x+1}{x\left(x^{2}+1\right)}\right] d x \\ &I=\int d x-\int \frac{x+1}{x\left(x^{2}+1\right)} d x \end{aligned}$
$\begin{aligned} &\frac{x+1}{x\left(x^{2}+1\right)}=\frac{A}{x}+\frac{B x+C}{x^{2}+1} \\ &x+1=A\left(x^{2}+1\right)+(B x+C) x \\ &x+1=x^{2}(A+B)+C x+A \end{aligned}$
Equating similar terms
$\begin{aligned} &A+B=0 \\ &A=-B \end{aligned}$ (1)
$C=1$ (2)
$\begin{aligned} &A=1 \\ &B=-1 \end{aligned}$ [From equation (1)]
$\frac{x+1}{x\left(x^{2}+1\right)}=\frac{1}{x}+\frac{-x+1}{x^{2}+1}$
Thus
$\begin{aligned} &I=\int d x-\int \frac{1}{x} d x+\int \frac{x}{x^{2}+1} d x-\int \frac{1}{x^{2}+1} d x \\ &I=x-\log |x|+\frac{1}{2} \log \left|x^{2}+1\right|-\tan ^{-1} x+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 45

Answer:
$-2 \log |x+1|-\frac{1}{x+1}+3 \log |x+2|+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{x^{2}+x+1}{(x+1)^{2}(x+2)} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{x^{2}+x+1}{(x+1)^{2}(x+2)} d x \\ &\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}=\frac{A}{x+1}+\frac{B}{(x+1)^{2}}+\frac{C}{x+2} \\ &x^{2}+x+1=A(x+1)(x+2)+B(x+2)+C(x+1)^{2} \end{aligned}$ …..(1)
For $x=-1$ equation (1) becomes
$\begin{aligned} &1-1+1=A(0)+B(1)+C(0) \\ &1=B \end{aligned}$
For $x=-2$ equation (1) becomes
$\begin{aligned} &4-2+1=A(0)+B(0)+C(1) \\ &C=3 \end{aligned}$
For $x=0$ equation (1) becomes
$\begin{aligned} &0+0+1=A(2)+B(2)+C \\ &1=2 A+2+3 \end{aligned}$ [As $B=1,C= 3$ ]
$\begin{aligned} &1=2 A+2+3 \\ &1=2 A+5 \\ &2 A=-4 \\ &A=-2 \\ &\therefore \frac{x^{2}+x+1}{(x+1)^{2}(x+2)}=\frac{-2}{x+1}+\frac{1}{(x+1)^{2}}+\frac{3}{x+2} \end{aligned}$
Thus
$\begin{aligned} &I=-2 \int \frac{d x}{x+1}+\int \frac{d x}{(x+1)^{2}}+3 \int \frac{d x}{x+2} \\ &I=-2 \log |x+1|-\frac{1}{x+1}+3 \log |x+2|+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 46

Answer:
$\frac{1}{4} \log \left|\frac{x^{4}}{x^{4}+1}\right|+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{1}{x\left(x^{4}+1\right)} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{1}{x\left(x^{4}+1\right)} d x \\ &I=\int \frac{x^{3}}{x^{4}\left(x^{4}+1\right)} d x \end{aligned}$ [Multiply and divide by $x^{3}$ ]
Let
$\begin{aligned} &x^{4}=y \\ &4 x^{3} d x=d y \\ &x^{3} d x=\frac{d y}{4} \\ &I=\frac{1}{4} \int \frac{d y}{y(y+1)} \end{aligned}$
Now
$\begin{aligned} &\frac{1}{y(y+1)}=\frac{A}{y}+\frac{B}{y+1} \\ &1=A(y+1)+B y \\ &1=(A+B) y+A \end{aligned}$
Equating similar terms
$\begin{aligned} &A=1 \\ &A+B=0 \\ &A=-B \\ &B=-1 \end{aligned}$
$\begin{aligned} &\frac{1}{y(y+1)}=\frac{1}{y}+\frac{(-1)}{y+1} \\ &I=\frac{1}{4} \int\left(\frac{1}{y}\right) d y+\frac{1}{4} \int \frac{-1}{y+1} d y \\ &I=\frac{1}{4} \log |y|+\left(\frac{-1}{4}\right) \log |y+1|+C \end{aligned}$
$\begin{aligned} &I=\frac{1}{4} \log \left|\frac{y}{y+1}\right|+C \\ &I=\frac{1}{4} \log \left|\frac{x^{4}}{x^{4}+1}\right|+C \quad\quad\quad\quad\quad\quad\left[y=x^{4}\right] \end{aligned}$

Indefinite Integrals exercise 18.30 question 47

Answer:
$\frac{1}{8} \log x-\frac{1}{24} \log \left(x^{3}+8\right)+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{1}{x\left(x^{3}+8\right)} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{1}{x\left(x^{3}+8\right)} d x \\ &I=\int \frac{x^{2}}{x^{3}\left(x^{3}+8\right)} d x \end{aligned}$ [Multiply and divide by $x^{2}$ ]
Let
$\begin{aligned} &x^{3}=y \\ &3 x^{2} d x=d y \\ &x^{2} d x=\frac{d y}{3} \\ &I=\frac{1}{3} \int \frac{d y}{y(y+8)} \\ &\frac{1}{y(y+8)}=\frac{A}{y}+\frac{B}{y+8} \\ &1=A(y+8)+B y \\ &1=(A+B) y+8 A \end{aligned}$
Equating both side
$\begin{aligned} &8 A=1 \\ &A=\frac{1}{8} \\ &A+B=0 \\ &B=-A=\frac{-1}{8} \\ &\frac{1}{y(y+8)}=\frac{1}{8 y}-\frac{1}{8(y+8)} \end{aligned}$
Thus
$\begin{aligned} &I=\frac{1}{3} \int\left(\frac{1}{8 y}-\frac{1}{8(y+8)}\right) d y \\ &I=\frac{1}{24} \int \frac{1}{y} d y-\frac{1}{24} \int \frac{1}{y+8} d y \end{aligned}$
$\begin{aligned} &I=\frac{1}{24} \log |y|-\frac{1}{24} \log |y+8|+C \\ &I=\frac{1}{24} \log \left|x^{3}\right|-\frac{1}{24} \log \left|x^{3}+8\right|+C \\ &I=\frac{1}{8} \log |x|-\frac{1}{24} \log \left|x^{3}+8\right|+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 48

Answer:
$\frac{3}{4}\left[\log \left|\frac{1+x^{2}}{(1-x)^{2}}\right|+2 \tan ^{-1} x+C\right]$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{3}{(1-x)\left(1+x^{2}\right)} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{3}{(1-x)\left(1+x^{2}\right)} d x \\ &\frac{3}{(1-x)\left(1+x^{2}\right)}=\frac{A}{1-x}+\frac{B x+C}{1+x^{2}} \\ &3=A\left(1+x^{2}\right)+(B x+C)(1-x) \\ &3=x^{2}(A-B)+(B-C) x+A+C \end{aligned}$
Equating the similar terms
$\begin{aligned} &A-B=0 \\ &A=B \end{aligned}$ (1)
$\begin{aligned} &B-C=0 \\ &B=C \end{aligned}$ (2)
$A+C= 3$
$B+B=3$ [From equation (1) and (2)]
$\begin{aligned} &2 B=3 \\ &B=\frac{3}{2} \\ &A=B=C=\frac{3}{2} \\ &\therefore \frac{3}{\left(1+x^{2}\right)(1-x)}=\frac{3}{2(1-x)}+\frac{3 x+3}{2\left(1+x^{2}\right)} \end{aligned}$
Thus
$\begin{aligned} &I=\frac{3}{2} \int \frac{1}{1-x} d x+\frac{3}{2} \int \frac{x}{1+x^{2}} d x+\frac{3}{2} \int \frac{1}{1+x^{2}} d x \\ &I=\frac{-3}{2} \log |1-x|+\frac{3}{4} \log \left|1+x^{2}\right|+\frac{3}{2} \tan ^{-1} x+C \\ &I=\frac{3}{4}\left[\log \left|\frac{1+x^{2}}{(1-x)^{2}}\right|+2 \tan ^{-1} x+C\right] \end{aligned}$

Indefinite Integrals exercise 18.30 question 49

Answer:
$\frac{-1}{27} \log |-\sin x+1|+\frac{1}{9(1-\sin x)}+\frac{1}{6(1-\sin x)^{2}}+\frac{1}{27} \log |2+\sin x|+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{\cos x d x}{(1-\sin x)^{3}(2+\sin x)}$
Explanation:
Let
$I=\int \frac{\cos x d x}{(1-\sin x)^{3}(2+\sin x)}$
Let
$\begin{aligned} &\sin x=y \\ &\cos x d x=d y \\ &I=\int \frac{d y}{(1-y)^{3}(2+y)} \end{aligned}$
Now
$\begin{aligned} &\frac{1}{(1-y)^{3}(2+y)}=\frac{A}{1-y}+\frac{B}{(1-y)^{2}}+\frac{C}{(1-y)^{3}}+\frac{D}{2+y} \\ &1=A(1-y)^{2}(2+y)+B(1-y)(2+y)+C(2+y)+D(1-y)^{3} \end{aligned}$
$\begin{aligned} &\text { Put } y=1 \\ &1=A(0)+B(0)+C(3)+D(0) \\ &1=3 C \end{aligned}$
$C= \frac{1}{3}$ (1)
$\begin{aligned} &\text { Put } y=-2 \\ &1=A(0)+B(0)+C(0)+D(27) \\ &1=27 D \\ &D=\frac{1}{27} \end{aligned}$

$\begin{aligned} &\text { Similarly } A=\frac{-1}{27}, B=\frac{1}{9} \\ &\frac{1}{(1-y)^{3}(2+y)}=\frac{-1}{27(1-y)}+\frac{1}{9(1-y)^{2}}+\frac{1}{3(1-y)^{3}}+\frac{1}{27(2+y)} \end{aligned}$

Thus
$I=\frac{-1}{27} \int \frac{1}{1-y} d y+\frac{1}{9} \int \frac{1}{(1-y)^{2}} d y+\frac{1}{3} \int \frac{1}{(1-y)^{3}} d y+\frac{1}{27} \int \frac{1}{2+y} d y$
$\begin{aligned} &I=\frac{-1}{27} \log |1-y|+\frac{1}{9(1-y)}+\frac{1}{6(1-y)^{2}}+\frac{1}{27} \log |2+y|+C \\ &\ldots \ldots \ldots .\left[\int \frac{1}{(1-a)^{2}} d a=\frac{1}{(1-a)}\right] \end{aligned}$
$I=\frac{-1}{27} \log |1-\sin x|+\frac{1}{9[1-\sin x]}+\frac{1}{6(1-\sin x)}+\frac{1}{27} \log |2+\sin x|+C$

Indefinite Integrals exercise 18.30 question 50

Answer:
$-\frac{1}{4 x}+\frac{7}{8} \tan ^{-1}\left(\frac{x}{2}\right)+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{2 x^{2}+1}{x^{2}\left(x^{2}+4\right)}dx$
Explanation:
Let
$I= \int \frac{2 x^{2}+1}{x^{2}\left(x^{2}+4\right)}dx$
Now let’s separate the fraction $\frac{2 x^{2}+1}{x^{2}\left(x^{2}+4\right)}$ through the partial fraction
Put
$\begin{aligned} &4 A=1 \\ &A=\frac{1}{4} \\ &A+B=2 \\ &B=2-A \\ &B=2-A \\ &B=2-\frac{1}{4} \\ &B=\frac{7}{4} \end{aligned}$
Equating both side
$\begin{aligned} &\frac{2 y+1}{y(y+4)}=\frac{1}{4 y}+\frac{7}{4(y+4)} \\ &=\frac{1}{4 x^{2}}+\frac{7}{4\left(x^{2}+4\right)} \end{aligned}$
$\begin{aligned} &I=\frac{1}{4} \int \frac{d x}{x^{2}}+\frac{7}{4} \int \frac{1}{x^{2}+4} d x \\ &I=\frac{1}{4} \cdot\left(\frac{1}{-x}\right)+\frac{7}{4} \cdot\left(\frac{1}{2}\right) \tan ^{-1} \frac{x}{2}+C \\ &I=\frac{-1}{4 x}+\frac{7}{8} \tan ^{-1} \frac{x}{2}+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 51

Answer:
$\log \left|\frac{2-\sin x}{1-\sin x}\right|+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{\cos x}{(1-\sin x)(2-\sin x)} d x$
Explanation:
Let
$I=\int \frac{\cos x}{(1-\sin x)(2-\sin x)} d x$
Let
$\begin{aligned} &\sin x=y \\ &\cos x d x=d y \\ &I=\int \frac{d y}{(1-y)(2-y)} \\ &\frac{1}{(1-y)(2-y)}=\frac{A}{1-y}+\frac{B}{2-y} \\ &1=A(2-y)+B(1-y) \end{aligned}$
Put $y= 2$
$\begin{aligned} &1=A(0)+B(-1) \\ &B=-1 \end{aligned}$
Put
$\begin{aligned} &y=1 \\ &1=A(1)+B(0) \\ &A=1 \end{aligned}$
$\begin{aligned} &\frac{1}{(2-y)(1-y)}=\frac{-1}{2-y}+\frac{1}{1-y} \\ &I=\int \frac{1}{1-y} d y-\int \frac{1}{2-y} d y \end{aligned}$
$\begin{aligned} &I=-\log |1-y|+\log |2-y|+C \\ &I=\log \left|\frac{2-y}{1-y}\right|+C \\ &I=\log \left|\frac{2-\sin x}{1-\sin x}\right|+C \end{aligned}$ $\quad[y=\sin x]$

Indefinite Integrals exercise 18.30 question 52

Answer:
$\log \left|\frac{(x-3)^{7}}{(x-2)^{5}}\right|+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{2 x+1}{(x-2)(x-3)} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{2 x+1}{(x-2)(x-3)} d x \\ &\frac{2 x+1}{(x-2)(x-3)}=\frac{A}{x-2}+\frac{B}{x-3} \end{aligned}$
$(2 x+1)=A(x-3)+B(x-2)$ (1)
Put $x= 3$
Equation (1)
$\begin{aligned} &6+1=B \\ &B=7 \end{aligned}$
Put $x= 2$
Equation (1)
$\begin{aligned} &4+1=A(-1) \\ &A=-5 \\ &\frac{2 x+1}{(x-2)(x-3)}=\frac{-5}{x-2}+\frac{7}{x-3} \end{aligned}$
$\begin{aligned} &I=-5 \int \frac{d x}{x-2}+7 \int \frac{d x}{x-3} \\ &I=-5 \log |x-2|+7 \log |x-3|+C \\ &I=\log \left|\frac{(x-3)^{7}}{(x-2)^{5}}\right|+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 53

Answer:
$\tan ^{-1} x-\frac{1}{\sqrt{2}} \tan ^{-1} \frac{x}{\sqrt{2}}+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{1}{\left(x^{2}+1\right)\left(x^{2}+2\right)} d x$
Explanation:
Let
$I=\int \frac{1}{\left(x^{2}+1\right)\left(x^{2}+2\right)} d x$
Let $x^{2}= y$
$\begin{aligned} &\frac{1}{\left(x^{2}+1\right)\left(x^{2}+2\right)}=\frac{1}{(y+1)(y+2)}=\frac{A}{y+1}+\frac{B}{y+2} \\ &\frac{1}{(y+1)(y+2)}=\frac{(y+2) A+(y+1) B}{(y+1)(y+2)} \\ &1=(y+2) A+(y+1) B \end{aligned}$ (1)
Put $y= -1$
Equation (1)
$\begin{aligned} &1=A+0 \\ &A=1 \end{aligned}$
Put $y= -2$
Equation (1)
$\begin{aligned} &1=0+(-1) B \\ &B=-1 \\ &\frac{1}{(y+1)(y+2)}=\frac{1}{y+2}-\frac{1}{y+2} \\ &\frac{1}{\left(x^{2}+1\right)\left(x^{2}+2\right)}=\frac{1}{x^{2}+1}-\frac{1}{x^{2}+2} \end{aligned}$

$\begin{aligned} &I=\int \frac{1}{x^{2}+1} d x-\int \frac{1}{x^{2}+2} d x \\ &I=\tan ^{-1} x-\frac{1}{\sqrt{2}} \tan ^{-1} \frac{x}{\sqrt{2}}+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 54

Answer:
$\frac{1}{4} \log \left|\frac{x^{4}-1}{x^{4}}\right|+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{1}{x\left(x^{4}-1\right)} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{1}{x\left(x^{4}-1\right)} d x \\ &I=\int \frac{x^{3}}{x^{4}\left(x^{4}-1\right)} d x \end{aligned}$ [Multiply and divide by $x^{3}$ ]
Let $x^{4}=y$
$4x^{3}dx= dy$
$\begin{aligned} &I=\frac{1}{4} \int \frac{d y}{y(y-1)} \\ &I=\frac{1}{4} \int \frac{1+y-y}{y(y-1)} d y \\ &I=\frac{1}{4} \int \frac{y-(y-1)}{y(y-1)} d y \end{aligned}$
$\begin{aligned} &I=\frac{1}{4}\left[\int \frac{1}{y-1} d y+\int \frac{-1}{y} d y\right] \\ &I=\frac{1}{4}[\log |y-1|-\log |y|]+C \\ &I=\frac{1}{4} \log \left|\frac{y-1}{y}\right|+C \end{aligned}$
As
$\begin{aligned} &y=x^{4} \\ &I=\frac{1}{4} \log \left|\frac{x^{4}-1}{x^{4}}\right|+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 55

Answer:
$\frac{1}{4} \log \left|\frac{x-1}{x+1}\right|-\frac{1}{2} \tan ^{-1} x+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{1}{x^{4}-1} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{1}{x^{4}-1} d x \\ &I=\int \frac{1}{(x-1)(x+1)\left(x^{2}+1\right)} d x \\ &\frac{1}{(x-1)(x+1)\left(x^{2}+1\right)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C x+D}{x^{2}+1} \\ &1=A\left(x^{2}+1\right)(x+1)+B(x-1)\left(x^{2}+1\right)+(C x+D)\left(x^{2}-1\right) \end{aligned}$
Put $x=-1$
$\begin{aligned} &1=A(0)+B(-2)(2)+(C x+D)(0) \\ &1=-4 B \\ &B=\frac{-1}{4} \end{aligned}$
Put $x=1$
$\begin{aligned} &1=A(2)(2)+B(0)+(C+D)(0) \\ &1=4 A \\ &A=\frac{1}{4} \end{aligned}$
Put $x=0$
$\begin{aligned} &1=A-B+D(-1) \\ &1=\frac{1}{4}+\frac{1}{4}-D \\ &1=\frac{1}{2}-D \\ &D=\frac{-1}{2} \end{aligned}$
Put $x=2$
$\begin{aligned} &1=A(5)(3)+B(1)(5)+(5 C+D)(3) \\ &1=15 A+5 B+15 C+3 D \\ &1=\frac{15}{4}-\frac{5}{4}+15 C-\frac{3}{2} \\ &1=\frac{10}{4}-\frac{3}{2}+15 C \\ &1=\frac{10-6}{4}+15 C \end{aligned}$
$\begin{aligned} &1=\frac{4}{4}+15 C \\ &1=1+15 C \\ &C=0 \\ &\frac{1}{(x-1)(x+1)\left(x^{2}+1\right)}=\frac{1}{4(x-1)}-\frac{1}{4(x+1)}-\frac{1}{2\left(x^{2}+1\right)} \end{aligned}$
Thus
$\begin{aligned} &I=\frac{1}{4} \int \frac{1}{x-1} d x-\frac{1}{4} \int \frac{d x}{x+1}-\frac{1}{2} \int \frac{1}{x^{2}+1} d x \\ &I=\frac{1}{4} \log |x-1|-\frac{1}{4} \log |x+1|-\frac{1}{2} \tan ^{-1} x+C \\ &I=\frac{1}{4} \log \left|\frac{x-1}{x+1}\right|-\frac{1}{2} \tan ^{-1} x+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 56

Answer:
$\log \left|x^{2}+1\right|-\log \left|x^{2}+2\right|+\frac{1}{\left(x^{2}+2\right)}+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{2 x}{\left(x^{2}+1\right)\left(x^{2}+2\right)^{2}} d x$
Explanation:
Let
$I=\int \frac{2 x}{\left(x^{2}+1\right)\left(x^{2}+2\right)^{2}} d x$
Let
$\begin{aligned} &x^{2}=y \\ &2 x d x=d y \\ &I=\int \frac{d y}{(y+1)(y+2)^{2}} \\ &\frac{1}{(y+1)(y+2)^{2}}=\frac{A}{y+1}+\frac{B}{y+2}+\frac{C}{(y+2)^{2}} \\ &1=A(y+2)^{2}+B(y+2)(y+1)+C(y+1) \end{aligned}$

Put $y=-2$

$\begin{aligned} &1=A+(0)+(0) \\ &A=1 \end{aligned}$
Put $y=0$
$\begin{aligned} &1=4 A+2 B+C \\ &1=4+2 B-1 \\ &1=2 B+3 \\ &-2=2 B \\ &B=-1 \end{aligned}$

$\begin{aligned} &\frac{1}{(y+1)(y+2)^{2}}=\frac{1}{y+1}-\frac{1}{(y+2)}-\frac{1}{(y+2)^{2}} \\ &I=\int \frac{d y}{1+y}-\int \frac{d y}{y+2}-\int \frac{d y}{(y+2)^{2}} \\ &I=\log |1+y|-\log |y+2|+\frac{1}{(y+2)}+C \end{aligned}$

As $y=x^{2}$
$I=\log \left|x^{2}+1\right|-\log \left|x^{2}+2\right|+\frac{1}{\left(x^{2}+2\right)}+C$

Indefinite Integrals exercise 18.30 question 57

Answer:
$\frac{1}{2} \log |x-1|+\frac{1}{4} \log \left|x^{2}+1\right|+\frac{1}{2} \tan ^{-1} x+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{x^{2}}{(x-1)\left(x^{2}+1\right)} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{x^{2}}{(x-1)\left(x^{2}+1\right)} d x \\ &\frac{x^{2}}{(x-1)\left(x^{2}+1\right)}=\frac{A}{x-1}+\frac{B x+C}{x^{2}+1} \\ &x^{2}=A\left(x^{2}+1\right)+(B x+C) x-B x-C \\ &x^{2}=x^{2}(A+B)+(-B+C) x+A-C \end{aligned}$

Equating both side

$\begin{aligned} &A-C=0 \\ &A=C \end{aligned}$ (1)
$\begin{aligned} &-B+C=0 \\ &B=C \end{aligned}$ (2)
$\begin{aligned} &A+B=1 \\ &C+C=1 \\ &2 C=1 \\ &C=\frac{1}{2} \\ &A=B=C=\frac{1}{2} \end{aligned}$
$\begin{aligned} &\frac{x^{2}}{(x-1)\left(x^{2}+1\right)}=\frac{1}{2(x-1)}+\frac{x+1}{2\left(x^{2}+1\right)} \\ &I=\frac{1}{2} \int \frac{d x}{x-1}+\frac{1}{2} \int \frac{x d x}{x^{2}+1}+\frac{1}{2} \int \frac{1}{x^{2}+1} d x \\ &I=\frac{1}{2} \log |x-1|+\frac{1}{4} \log \left|x^{2}+1\right|+\frac{1}{2} \tan ^{-1} x+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 58

Answer:
$\frac{a}{a^{2}-b^{2}} \tan ^{-1} \frac{x}{a}+\frac{b}{b^{2}-a^{2}} \tan ^{-1} \frac{x}{b}+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{x^{2}}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)} d x$
Explanation:
Let
$I=\int \frac{x^{2}}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)} d x$
$\frac{x^{2}}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)}=\frac{A x+B}{\left(x^{2}+a^{2}\right)}+\frac{C x+D}{\left(x^{2}+b^{2}\right)} \\$ (i)
$x^{2}=(A x+B)\left(x^{2}+b^{2}\right)+(C x+D)\left(x^{2}+a^{2}\right)$
On comparing coefficient
$x^{2}$ Coefficient
$1=B+D$ (1)
$x$ Coefficient
$0=A b^{2}+C a^{2}$ (2)
Constant
$0=B b^{2}+D a^{2}$ (3)
$\begin{aligned} &x^{3} \text { Coefficient }\\ &0=A+C \end{aligned}$ (4)
Solving (2) and (4)
$\begin{aligned} &A b^{2}+C a^{2}=0 \\ &A+C=0 \\ &C=-A \\ &A b^{2}-A a^{2}=0 \\ &A\left(b^{2}-a^{2}\right)=0 \\ &A=0, C=0 \end{aligned}$
$B+D=1$ (1)
$B b^{2}+D a^{2}=0$ (3)
Multiply (1) by $b^{2}$ and subtract (3)
$\begin{aligned} &B b^{2}+D b^{2}=b^{2}- \\ &B b^{2}+D a^{2}=0 \\ &\overline{D\left(b^{2}-a^{2}\right)=b^{2}} \end{aligned}$
$\begin{aligned} &D=\frac{b^{2}}{b^{2}-a^{2}} \\ &B=1-D \\ &B=1-\frac{b^{2}}{b^{2}-a^{2}} \\ &B=\frac{-a^{2}}{b^{2}-a^{2}} \end{aligned}$
(i) Becomes
$\begin{aligned} &\frac{x^{2}}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)} d x=\frac{-a^{2}}{b^{2}-a^{2}} \int \frac{1}{x^{2}+a^{2}} d x+\frac{b^{2}}{b^{2}-a^{2}} \int \frac{1}{x^{2}+b^{2}} d x \\ &=\frac{a^{2}}{a^{2}-b^{2}} \cdot \frac{1}{a} \tan ^{-1} \frac{x}{a}+\frac{b^{2}}{b^{2}-a^{2}} \cdot \frac{1}{b} \tan ^{-1} \frac{x}{b} \\ &=\frac{a}{a^{2}-b^{2}} \tan ^{-1} \frac{x}{a}+\frac{b}{b^{2}-a^{2}} \tan ^{-1} \frac{x}{b}+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 59

Answer:
$\frac{1}{18} \log (1+\sin x)-\frac{1}{2} \log (1-\sin x)+\frac{4}{9} \log \left(\frac{5}{4}-\sin x\right)+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{1}{\cos x(5-4 \sin x)} d x$
Explanation:
Let
$I=\int \frac{1}{\cos x(5-4 \sin x)} d x$
$\begin{aligned} &\text { Put }\\ &\sin x=t\\ &\cos x d x=d t\\ &d x=\frac{1}{\cos x} d t \end{aligned}$

$\begin{aligned} &I=\int \frac{1}{\cos ^{2} x(5-4 t)}=\int \frac{1}{\left(1-\sin ^{2} x\right)(5-4 t)} d t \\ &I=\int \frac{1}{\left(1-t^{2}\right)(5-4 t)} d t \\ &I=\frac{1}{4} \int \frac{1}{(1+t)(1-t)\left(\frac{5}{4}-t\right)} d t \end{aligned}$ (1)

$\frac{1}{(1+t)(1-t)\left(\frac{5}{4}-t\right)}=\frac{A}{1+t}+\frac{B}{1-t}+\frac{C}{\frac{5}{4}-t}$
$\begin{aligned} &\frac{1}{(1+t)(1-t)\left(\frac{5}{4}-t\right)}=\frac{\left[A(1-t)\left(\frac{5}{4}-t\right)+B(1+t)\left(\frac{5}{4}-t\right)+C\left(1-t^{2}\right)\right]}{(1+t)(1-t)\left(\frac{5}{4}-t\right)} \\ &1=A(1-t)\left(\frac{5}{4}-t\right)+B(1+t)\left(\frac{5}{4}-t\right)+C\left(1-t^{2}\right) \end{aligned}$
$\begin{aligned} &\text { Put } t=1 \\ &1=2 B\left(\frac{1}{4}\right) \\ &B=2 \end{aligned}$

$\text { Put } t=-1 \\$

$1=A(2)\left(\frac{9}{4}\right)$
$A=\frac{2}{9}$
$\begin{aligned} &\text { Put } t=\frac{5}{4} \\ &1=C\left(1-\frac{25}{16}\right) \\ &1=\frac{-9}{16} C \end{aligned}$
$\begin{aligned} &C=\frac{-16}{9} \\ &I=\frac{1}{4} \int \frac{\frac{2}{9}}{1+t} d t+\frac{1}{4} \int \frac{2}{(1-t)} d t-\frac{1}{4} \times \frac{16}{9} \int \frac{1}{\frac{5}{4}-t} d t \\ &I=\frac{1}{18} \log (1+t)-\frac{1}{2} \log (1-t)+\frac{4}{9} \log \left(\frac{5}{4}-t\right)+C \\ &I=\frac{1}{18} \log (1+\sin x)-\frac{1}{2} \log (1-\sin x)+\frac{4}{9} \log \left(\frac{5}{4}-\sin x\right)+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 60

Answer:
$\frac{1}{10} \log (\cos x-1)-\frac{1}{2} \log (\cos +1)+\frac{2}{5} \log \left(\frac{3}{2}+\cos x\right)+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{1}{\sin x(3+2 \cos x)} d x$
Explanation:
Let
$I=\int \frac{1}{\sin x(3+2 \cos x)} d x$
$\begin{aligned} &\text { Put }\\ &\operatorname{cox}=t\\ &-\sin x d x=d t\\ &d x=\frac{-1}{\sin x} d t\\ &I=\int \frac{-1}{\sin ^{2} x(3+2 t)} d t=\int \frac{-1}{\left(1-\cos ^{2} x\right)(3+2 t)} d t\\ &I=\int \frac{1}{\left(t^{2}-1\right)(3+2 t)} d t \end{aligned}$
$I=\frac{1}{2} \int \frac{1}{(t-1)(t+1)\left(\frac{3}{2}+t\right)} d t$ (1)
$\frac{1}{(t-1)(t+1)\left(\frac{3}{2}+t\right)}=\frac{A}{t-1}+\frac{B}{t+1}+\frac{C}{\frac{3}{2}+t}$
$\begin{aligned} &\frac{1}{(t-1)(t+1)\left(\frac{3}{2}+t\right)}=\frac{A(t+1)\left(\frac{3}{2}+t\right)+B(t-1)\left(\frac{3}{2}+t\right)+C\left(t^{2}-1\right)}{(t-1)(t+1)\left(\frac{3}{2}+t\right)} \\ &1=A(t+1)\left(\frac{3}{2}+t\right)+B(t-1)\left(\frac{3}{2}+t\right)+C\left(t^{2}-1\right) \end{aligned}$
$\begin{aligned} &\text { At } t=-1 \\ &1=-2 B\left(\frac{1}{2}\right) \\ &B=-1 \\ &\text { At } t=1 \\ &1=2 A\left(\frac{5}{2}\right) \end{aligned}$
$\begin{aligned} &A=\frac{1}{5} \\ &\text { At } t=\frac{-3}{2} \\ &1=C\left(\frac{9}{4}-1\right) \\ &C=\frac{4}{5} \end{aligned}$
Put in (1)
$\begin{aligned} &I=\frac{1}{2} \times \frac{1}{5} \int \frac{1}{t-1} d t+\frac{(-1)}{2} \int \frac{1}{t+1} d t+\frac{1}{2} \times \frac{4}{5} \int \frac{1}{\frac{3}{2}+t} d t \\ &I=\frac{1}{10} \log (t-1)-\frac{1}{2} \log (t+1)+\frac{2}{5} \log \left(\frac{3}{2}+t\right)+C \\ &I=\log (\cos x-1)-\frac{1}{2} \log (\cos x+1)+\frac{2}{5} \log \left(\frac{3}{2}+\cos x\right)+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 61

Answer:
$\frac{1}{6} \log (\cos x-1)+\frac{1}{2} \log (\cos x+1)-\frac{2}{3} \log \left(\cos x+\frac{1}{2}\right)+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{1}{\sin x+\sin 2 x} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{1}{\sin x+\sin 2 x} d x \\ &I=\int \frac{1}{\sin x+2 \sin x \cos x} d x \\ &I=\int \frac{1}{\sin x(1+2 \cos x)} d x \end{aligned}$
Put
$\begin{aligned} &\cos x=t \\ &-\sin x d x=d t \\ &d x=\frac{-1}{\sin x} d x \\ &I=\int \frac{-1}{\sin ^{2} x(1+2 t)} d t \\ &I=\int \frac{-1}{\left(1-\cos ^{2} x\right)(1+2 t)} d t \end{aligned}$
$I=\int \frac{1}{2\left(t^{2}-1\right)\left(\frac{1}{2}+t\right)} d t$ (1)
$\frac{1}{(t-1)(t+1)\left(\frac{1}{2}+t\right)}=\frac{A}{t-1}+\frac{B}{t+1}+\frac{C}{\frac{1}{2}+t}$ (2)
$1=A(t+1)\left(\frac{1}{2}+t\right)+B(t-1)\left(\frac{1}{2}+t\right)+C\left(t^{2}-1\right)$
At $t=- 1$
$\begin{aligned} &1=-2 B\left(\frac{-1}{2}\right) \\ &B=1 \end{aligned}$
At $t= 1$
$\begin{aligned} &1=2 A\left(\frac{3}{2}\right) \\ &A=\frac{1}{3} \end{aligned}$
$\begin{aligned} &\text { At } t=-\frac{1}{2} \\ &1=C\left(\frac{1}{4}-1\right) \\ &1=C\left(\frac{-3}{4}\right) \\ &C=\frac{-4}{3} \end{aligned}$
Put in (1) using (2)
$\begin{aligned} &I=\frac{1}{2} \cdot \frac{1}{3} \int \frac{1}{t-1} d t+\frac{1}{2} \int \frac{1}{t+1} d t-\frac{1}{2} \cdot \frac{4}{3} \int \frac{1}{t+\frac{1}{2}} d t \\ &I=\frac{1}{6} \log (t-1)+\frac{1}{2} \log (t+1)-\frac{2}{3} \log \left(t+\frac{1}{2}\right)+C \\ &I=\frac{1}{6} \log (\cos x-1)+\frac{1}{2} \log (\cos x+1)-\frac{2}{3} \log \left(\cos x+\frac{1}{2}\right)+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 62

Answer:
$\log \frac{x e^{x}}{1+x e^{x}}+c$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{x+1}{x\left(1+x e^{x}\right)} d x$
Explanation:
Let
$I=\int \frac{x+1}{x\left(1+x e^{x}\right)} d x$
Put
$1+x e^{x}=t$
Differentiate w.r.t $x$
$\begin{aligned} &x e^{x}+e^{x} d x=d t \\ &e^{x}(x+1) d x=d t \\ &(x+1) d x=\frac{1}{e^{x}} d t \\ &I=\int \frac{1}{x e^{x}(t)} d t \\ &I=\int \frac{1}{(t-1) t} d t \end{aligned}$ (1)
Let
$\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{t-1}$ (2)
$\begin{aligned} &1=A(t-1)+B(t) \\ &\text { At } t=1 \\ &1=B \\ &\text { At } t=0 \\ &\begin{array}{l} 1=-A \\ A=-1 \end{array} \end{aligned}$
Put in (1) using (2)
$\begin{aligned} &I=\int \frac{-1}{t} d t+\int \frac{1}{t-1} d t \\ &I=-\log t+\log (t-1)+C \\ &I=\log \frac{t-1}{t}+C \\ &I=\log \frac{x e^{x}}{1+x e^{x}}+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 63

Answer:
$x+\frac{2}{3} \tan ^{-1} \frac{x}{\sqrt{3}}-3 \tan ^{-1} \frac{x}{2}$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{\left(x^{2}+1\right)\left(x^{2}+2\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{\left(x^{2}+1\right)\left(x^{2}+2\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)} d x \\ &I=\int \frac{x^{4}+3 x^{2}+2}{x^{4}+7 x^{2}+12} d x \\ &I=\int\left(1-\frac{\left(4 x^{2}+10\right)}{x^{4}+7 x^{2}+12}\right) d x \\ &I=\int 1 d x-\int \frac{4 x^{2}+10}{x^{4}+7 x^{2}+12} d x \\ &I=x-I_{1} \end{aligned}$ (1)
Where
$\begin{aligned} &I_{1}=\int \frac{4 x^{2}+10}{x^{4}+7 x^{2}+12} d x \\ &\frac{4 x^{2}+10}{x^{4}+7 x^{2}+12}=\frac{4 x^{2}+10}{\left(x^{2}+3\right)\left(x^{2}+4\right)} \end{aligned}$ (i)
$4 x^{2}+10=(A x+B)\left(x^{2}+4\right)+(C x+D)\left(x^{2}+3\right)$
Comparing the coefficient
$x^{3}$ Coefficient
$0=A+C$ (2)
$x^{2}$ Coefficient
$4=B+D$ (3)
$x$ Coefficient
$0=4 A+3 C$ (4)
Constant
$10=4 B+3 D$ (5)
$\begin{aligned} &\text { Term (2) } A=-C \text { put in (4) }\\ &0=-4 C+3 C\\ &C=0\\ &A=0 \end{aligned}$
Multiply (3) by 4 and subtract it from (5)
$\begin{aligned} &\begin{array}{l} 4 B+4 D=16 \\ 4 B+3 D=10 \\ \hline D=6 \\ B+D=4 \\ B+6=4 \\ B=-2 \end{array} \end{aligned}$
(i) Becomes
$\begin{aligned} &I_{1}=\int \frac{-2}{x^{2}+3} d x+6 \int \frac{1}{x^{2}+4} d x \\ &I_{1}=-2 \cdot \frac{1}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}+6 \cdot \frac{1}{2} \tan ^{-1} \frac{x}{2} \end{aligned}$
$\begin{aligned} &I_{1}=\frac{-2}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}+3 \tan ^{-1} \frac{x}{2} \\ &I=x-I_{1} \\ &I=x+\frac{2}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}-3 \tan ^{-1} \frac{x}{2}+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 64

Answer:
$\frac{19}{2 \sqrt{2}} \tan ^{-1} \frac{x}{\sqrt{2}}-\frac{39}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}+\frac{67}{4} \tan ^{-1} \frac{x}{2}+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{4 x^{4}+3}{\left(x^{2}+2\right)\left(x^{2}+3\right)\left(x^{2}+4\right)} d x$
Explanation:
Let
$I=\int \frac{4 x^{4}+3}{\left(x^{2}+2\right)\left(x^{2}+3\right)\left(x^{2}+4\right)} d x$
Put $x^{2}= y$
$I=\int \frac{4 y^{2}+3}{(y+2)(y+3)(y+4)}$
Let
$\begin{aligned} &\frac{4 y^{2}+3}{(y+2)(y+3)(y+4)}=\frac{A}{y+2}+\frac{B}{y+3}+\frac{C}{y+4} \\ &4 y^{2}+3=A(y+3)(y+4)+B(y+2)(y+4)+C(y+2)(y+3) \end{aligned}$
$\begin{aligned} &y=-3 \\ &39=-B \\ &y=-4 \\ &67=C(-2)(-1) \\ &C=\frac{67}{2} \\ &y=-2 \\ &19=2 A \\ &A=\frac{19}{2} \end{aligned}$
$\begin{aligned} &I=\frac{19}{2} \int \frac{d x}{x^{2}+2}-39 \int \frac{d x}{x^{2}+3}+\frac{67}{2} \int \frac{d x}{x^{2}+4} \\ &I=\frac{19}{2 \sqrt{2}} \tan ^{-1} \frac{x}{\sqrt{2}}-\frac{39}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}+\frac{67}{4} \tan ^{-1} \frac{x}{2}+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 65

Answer:
$\frac{x^{2}}{2}+x+\frac{1}{2} \log |x-1|-\frac{1}{4} \log \left|x^{2}+1\right|-\frac{1}{2} \tan ^{-1} x+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{x^{4}}{(x-1)\left(x^{2}+1\right)} d x$
Explanation:
Let
$\begin{aligned} &I=\int \frac{x^{4}}{(x-1)\left(x^{2}+1\right)} d x \\ &I=\int \frac{x^{4}}{x^{3}-x^{2}+x-1} d x \\ &I=\frac{x\left(x^{3}-x^{2}+x-1\right)+1\left(x^{3}-x^{2}+x-1\right)+1}{\left(x^{3}-x^{2}+x-1\right)} \\ &I=x+1+\frac{1}{(x-1)\left(x^{2}+1\right)} \end{aligned}$
Let,
$\begin{aligned} &\frac{1}{(x-1)\left(x^{2}+1\right)}=\frac{A}{x-1}+\frac{B x+C}{x^{2}+1} \\ &1=A\left(x^{2}+1\right)+(B x+C)(x-1) \end{aligned}$
Put $x=1$
$\begin{aligned} &1=2 A \\ &A=\frac{1}{2} \end{aligned}$
Put $x= 0$
$\begin{aligned} &1=A-C \\ &C=A-1=-\frac{1}{2} \\ &C=\frac{-1}{2} \end{aligned}$
Put $x= -1$
$\begin{aligned} &1=2 A+2 B-2 C \\ &1=2(A-C)+2 B \\ &1=2+2 B \\ &2 B=-1 \\ &B=\frac{-1}{2} \end{aligned}$
$\begin{aligned} &\int \frac{x^{2}}{(x-1)\left(x^{2}+1\right)} d x=\int x d x+\int 1 d x+\frac{1}{2} \int \frac{1}{x-1} d x-\frac{1}{2} \int \frac{x+1}{x^{2}+1} d x \\ &=\frac{x^{2}}{2}+x+\frac{1}{2} \log |x-1|-\frac{1}{4} \log \left|x^{2}+1\right|-\frac{1}{2} \tan ^{-1} x+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 66

Answer:
$\frac{1}{7} \log \frac{x-2}{x+2}+\frac{\sqrt{3}}{7} \tan ^{-1} \frac{x}{\sqrt{3}}+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{x^{2}}{x^{4}-x^{2}-12} d x$
Explanation:
Let
$I=\int \frac{x^{2}}{x^{4}-x^{2}-12} d x$
Let $x^{2}= y$
$\begin{aligned} &\frac{y}{y^{2}-y-12}=\frac{y}{(y-4)(y+3)} \\ &\frac{y}{(y-4)(y+3)}=\frac{A}{y-4}+\frac{B}{y+3} \\ &y=A(y+3)+B(y-4) \end{aligned}$
$\begin{aligned} &y=-3 \\ &-3=-7 B \\ &B=\frac{3}{7} \\ &y=4 \\ &4=7 A \\ &A=\frac{4}{7} \end{aligned}$
$\begin{aligned} &\int \frac{x^{2}}{x^{4}-x^{2}-12}=\frac{4}{7} \int \frac{1}{x^{2}-4} d x+\frac{3}{7} \int \frac{1}{x^{2}+3} d x \\ &=\frac{1}{7} \log \frac{x-2}{x+2}+\frac{3}{7} \cdot \frac{1}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 67

Answer:
$\frac{-1}{2} \tan ^{-1} x+\frac{1}{4} \log \left(\frac{1+x}{1-x}\right)+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{x^{2}}{1-x^{4}} d x$
Explanation:
$\begin{aligned} &\text { Let }\\ &I=\int \frac{x^{2}}{1-x^{4}} d x \end{aligned}$

$\begin{aligned} &\text { Let } x^{2}=y \\ &\frac{y}{1-y^{2}}=\frac{y}{(1+y)(1-y)}=\frac{A}{1+y}+\frac{B}{1-y} \\ &y=A(1-y)+B(1+y) \end{aligned}$
$\begin{gathered} \text { At } y=1 \\ 1=2 B \\ B=\frac{1}{2} \end{gathered}$
$\begin{aligned} &\text { At } y=-1 \\ &-1=2 A \\ &A=\frac{-1}{2} \\ &I=\frac{-1}{2} \int \frac{1}{x^{2}+1} d x+\frac{1}{2} \int \frac{1}{1-x^{2}} d x \\ &I=\frac{-1}{2} \tan ^{-1} x+\frac{1}{2} \int \frac{1}{1-x^{2}} d x \end{aligned}$
$\begin{aligned} &I=\frac{-1}{2} \tan ^{-1} x+\frac{1}{2} \int \frac{1}{(1+x)(1-x)} d x \\ &I=\frac{-1}{2} \tan ^{-1} x+\frac{1}{2} \cdot \frac{1}{2} \int \frac{2+x-x}{(1+x)(1-x)} d x \\ &I=\frac{-1}{2} \tan ^{-1} x+\frac{1}{4} \int \frac{(1+x)+(1-x)}{(1+x)(1-x)} d x \end{aligned}$
$\begin{aligned} &I=\frac{-1}{2} \tan ^{-1} x+\frac{1}{4} \int \frac{1}{1-x} d x+\frac{1}{4} \int \frac{1}{1+x} d x \\ &I=\frac{-1}{2} \tan ^{-1} x-\frac{1}{4} \log (1-x)+\frac{1}{4} \log (1+x)+C \\ &I=\frac{-1}{2} \tan ^{-1} x+\frac{1}{4} \log \left(\frac{1+x}{1-x}\right)+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 68

Answer:
$\frac{1}{6} \log \frac{x-1}{x+1}+\frac{\sqrt{2}}{3} \tan ^{-1} \frac{x}{\sqrt{2}}+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{x^{2}}{x^{4}+x^{2}-2} d x$
Explanation:
$\begin{aligned} &\text { Let }\\ &I=\int \frac{x^{2}}{x^{4}+x^{2}-2} d x \end{aligned}$
$\begin{aligned} &\text { Put } x^{2}=y \\ &\frac{y}{y^{2}+y-2}=\frac{y}{(y-1)(y+2)}=\frac{A}{y-1}+\frac{B}{y+2} \\ &y=A(y+2)+B(y-1) \end{aligned}$

$\! \! \! \! \! \! \! \! \! \! \text { At } y=1 \\ \\1=3 A \\\\ A=\frac{1}{3}$
$\begin{aligned} &\text { At } y=-2 \\ &-2=-3 B \\ &B=\frac{2}{3} \end{aligned}$
$\begin{aligned} &\int \frac{x^{2}}{x^{4}+x^{2}-2}=\frac{1}{3} \int \frac{1}{x^{2}-1} d x+\frac{2}{3} \int \frac{1}{x^{2}+2} d x \\ &=\frac{1}{3} \cdot \frac{1}{2} \log \frac{x-1}{x+1}+\frac{2}{3} \cdot \frac{1}{\sqrt{2}} \tan ^{-1} \frac{x}{\sqrt{2}}+C \\ &I=\frac{1}{6} \log \frac{x-1}{x+1}+\frac{\sqrt{2}}{3} \tan ^{-1} \frac{x}{\sqrt{2}}+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 69

Answer:
$\frac{1}{4 \sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}+\frac{27}{40} \log \frac{x-5}{x+5}+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{\left(x^{2}+1\right)\left(x^{2}+4\right)}{\left(x^{2}+3\right)\left(x^{2}-5\right)} d x \\$
Explanation:
Let
$\begin{aligned} &I=\int \frac{\left(x^{2}+1\right)\left(x^{2}+4\right)}{\left(x^{2}+3\right)\left(x^{2}-5\right)} d x \\ &I=\int \frac{x^{4}+5 x^{2}+4}{x^{4}-2 x^{2}-15} d x \\ &I=\int\left(1+\frac{7 x^{2}+19}{x^{4}-2 x^{2}-15}\right) d x \\ &I=x-\int \frac{7 x^{2}+19}{x^{4}-2 x^{2}-15} d x \end{aligned}$
$I=x-I_{1}$ (1)
Where
$\begin{aligned} &I_{1}=\int \frac{7 x^{2}+19}{x^{4}-2 x^{2}-15} d x \\ &x^{2}=y \\ &\frac{7 y+19}{y^{2}-2 y-15}=\frac{7 y+19}{(y+3)(y-5)}=\frac{A}{y+3}+\frac{B}{y-5} \\ &7 y+19=A(y-5)+B(y+3) \end{aligned}$
$\begin{aligned} &\text { At } y=5 \\ &54=8 B \\ &B=\frac{27}{4} \end{aligned}$
$\begin{aligned} &\text { At } y=-3 \\ &-2=-8 A \\ &A=\frac{1}{4} \end{aligned}$
$\begin{aligned} &\int \frac{7 x^{2}+19}{x^{4}-2 x^{2}-15}=\frac{1}{4} \int \frac{1}{x^{2}+3} d x+\frac{27}{4} \int \frac{1}{x^{2}-5} d x \\ &=\frac{1}{4} \cdot \frac{1}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}+\frac{27}{4} \cdot \frac{1}{2 \times 5} \log \frac{x-5}{x+5}+C \\ &=\frac{1}{4 \sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}+\frac{27}{40} \log \frac{x-5}{x+5}+C \end{aligned}$

Indefinite Integrals exercise 18.30 question 70

Answer:
$-\log (1-\sin x)+\frac{1}{2} \log \left(1+\sin ^{2} x\right)+\tan ^{-1}(\sin x)+C$
Hint:
To solve this integration, we use partial fraction method
Given:
$\int \frac{2 \cos x}{(1-\sin x)\left(1+\sin ^{2} x\right)} d x$
Explanation:
Let
$I=\int \frac{2 \cos x}{(1-\sin x)\left(1+\sin ^{2} x\right)} d x$
$\begin{aligned} &\text { Put } \sin x=t \\ &\cos x d x=d t \\ &I=\int \frac{2 d t}{(1-t)\left(1+t^{2}\right)} \end{aligned}$ (1)
$\begin{aligned} &\frac{2}{(1-t)\left(1+t^{2}\right)}=\frac{A}{1-t}+\frac{B t+C}{1+t^{2}} \\ &2=A\left(t^{2}+1\right)+(B t+C)(1-t) \\ &\text { At } t=1 \\ &\begin{array}{l} 2=2 A+(B+C)(0) \\ A=1 \end{array} \\ &\begin{array}{l} \end{array} \end{aligned}$
$\begin{aligned} &\text { At } t=0 \\ &\begin{array}{l} 2=A+C \\ 2=1+C \\ C=1 \\ \text { At } t=-1 \\ 2=2 A+(-B+C)(2) \\ 2=2(1)+(-B+1)(2) \\ 2=2+2(1-B) \\ 1-B=0 \\ B=1 \end{array} \end{aligned}$
$\begin{aligned} &I=\int \frac{1}{1-t} d t+\int \frac{t+1}{t^{2}+1} d t \\ &I=-\log (1-t)+\frac{1}{2} \int \frac{2 t}{t^{2}+1} d t+\int \frac{1}{t^{2}+1} d t \end{aligned}$ $\begin{aligned} &I=-\log (1-t)+\frac{1}{2} \log \left(1+t^{2}\right)+\tan ^{-1} t+C \\ &I=-\log (1-\sin x)+\frac{1}{2} \log \left(1+\sin ^{2} x\right)+\tan ^{-1}(\sin x)+C \end{aligned}$

The topics in chapter 18, ex 30 includes assessing the integrals using mathematical replacements, sums regarding coordinating and its methods, joining the parts and so on. Other than the sums provided in the textbook, there are various other questions and solved answers in the RD Sharma Class 12 Solutions Chapter 18 Exercise 18.30 Indefinite Integrals reference book.

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