RD Sharma Class 12 Exercise 18.19 Indefinite Integrals Solutions Maths - Download PDF Free Online

# RD Sharma Class 12 Exercise 18.19 Indefinite Integrals Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 12:20 PM IST

RD Sharma Class 12th Exercise 18.19 solutions are the best choice for a student because of the perfect answers to all the questions of RD Sharma which is considered as maths bible. The team always looks forward to finding multiple ways to solve these questions making the journey easy for every student. This particular exercise has 18 questions that are formatted by a team of experts. RD Sharma Solutions Rd Sharma Class 12th Exercise 18.19 discusses completing squaring method to solve quadratic equation, integrating the function, which includes use of fundamental integration formulae to explain standard results on integration.

## Indefinite Integrals Excercise:18.19

Indefinite Integrals Exercise 18.19 Question 1

Answer:$=\frac{1}{2}\left[\log \left|x^{2}+3 x+2\right|\right]-\frac{3}{2}\left[\log \left|\frac{x+1}{x+2}\right|\right]+c$
Hint Using Integration method
Given:$\int \frac{x}{x^{2}+3 x+2} d x$
Solution: Let,$x^{2}+3 x+2=t$
$(2 x+3) d x=d t$
\begin{aligned} &I=\frac{2}{2} \int \frac{x}{x^{2}+3 x+2} d x \\ &=\frac{1}{2} \int \frac{2 x}{x^{2}+3 x+2} d x \\ &=\frac{1}{2} \int \frac{2 x+3-3}{x^{2}+3 x+2} d x \\ &=\frac{1}{2}\left[\int \frac{2 x+3}{x^{2}+3 x+2} d x-\int \frac{3}{x^{2}+3 x+2} d x\right] \end{aligned}
$=\frac{1}{2}\left[\frac{d t}{t}-3 \int \frac{d x}{(x+1)(x+2)}\right]$ $\left[\begin{array}{c} \because\left(x^{2}+3 x+2\right) \\ =\left(x^{2}+2 x+x+2\right) \\ =(x(x+2)+1(x+2)) \\ =(x+1)(x+2) \end{array}\right]$
$=\frac{1}{2}\left[\log t-3 \int\left(\frac{1}{x+1}-\frac{1}{x+2}\right) d x\right]$ $\left[\int \frac{1}{x} d x=\log x+c\right]$
$=\frac{1}{2}\left[\log \left(x^{2}+3 x+2\right)-3(\log (x+1)-\log (x+2)]+c\right.$
$=\frac{1}{2}\left[\log \left(x^{2}+3 x+2\right)-3 \log \left(\frac{x+1}{x+2}\right)\right]+c$
$=\frac{1}{2}\left[\log \left(x^{2}+3 x+2\right)\right]-\frac{3}{2}\left[\log \left(\frac{x+1}{x+2}\right)\right]+c$

Indefinite Integrals Exercise 18.18 Question 2

Answer: $\frac{1}{2} \log \left|x^{2}+x+3\right|+\frac{1}{\sqrt{11}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{11}}\right)+c$
Hint: Integrate by ILATE
Given: $\int \frac{x+1}{x^{2}+x+3} d x$
Solution:
\begin{aligned} &I_{1}=\frac{1}{2} \int \frac{2 x+2}{x^{2}+x+3} d x \\ &I_{1}=\frac{1}{2} \int \frac{(2 x+1)+1}{x^{2}+x+3} d x \end{aligned}
$=\frac{1}{2} \int \frac{2 x+1}{x^{2}+x+3} d x+\frac{1}{2} \int \frac{1}{x^{2}+x+3} d x$
\begin{aligned} I_{1} &=\frac{1}{2} \int \frac{2 x+2}{x^{2}+x+3} d x \\ I_{1} &=\frac{1}{2} \int \frac{(2 x+1)+1}{x^{2}+x+3} d x \\ &=\frac{1}{2} \int \frac{2 x+1}{x^{2}+x+3} d x+\frac{1}{2} \int \frac{1}{x^{2}+x+3} d x \end{aligned} $\left[\int \frac{d t}{t}=\ln t+c,\right]$
Let,$x^{2}+x+3=t$
$(2 x+1) d x=d t$
$I_{1}=\log \left|x^{2}+x+3\right|+\mathrm{C}$
$I_{2}=\int \frac{1}{x^{2}+x+3} d x$
$=\int \frac{1}{\left(x+\frac{1}{2}\right)^{2}-\frac{1}{4}+3} d x$
$=\int \frac{1}{\left(x+\frac{1}{2}\right)^{2}+\frac{11}{4}} d x$ $\left[\begin{array}{c} x+\frac{1}{2}=t \Rightarrow d x=d t \\ \sqrt{\frac{11}{4}}=a=\frac{\sqrt{11}}{2} \end{array}\right]$
$=\frac{1}{\frac{\sqrt{11}}{2}} \tan ^{-1}\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{11}}{2}}\right)+C_{2}$ $\left[\int \frac{1}{t^{2}+a^{2}} d t=\frac{1}{a} \tan ^{-1}\left(\frac{t}{a}\right)+C_{2}\right]$
$I_{2}=\frac{2}{\sqrt{11}} \tan ^{-1} \frac{(2 x+1)}{\sqrt{11}}+C_{2}$
$I=\frac{1}{2}\left(I_{1}+I_{2}\right)$
$=\frac{1}{2}\left[\log \left|x^{2}+x+3\right|+\frac{2}{\sqrt{11}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{11}}\right)\right]+c$
$=\frac{1}{2} \log \left|x^{2}+x+3\right|+\frac{1}{\sqrt{11}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{11}}\right)+c$

Indefinite Integrals Exercise 18.19 Question 3

Answer: $-\frac{2}{\sqrt{5}} \log \left|\frac{x+1-\sqrt{5}}{x+1+\sqrt{5}}\right|+\frac{1}{2} \log \left|x^{2}+2 x-4\right|+c$
Hint Find value of A and B
Given: $\int \frac{x-3}{x^{2}+2 x-4} d x$
Solution: Let $x-3=A+B \frac{d}{d x}\left(x^{2}+2 x-4\right)$
$x-3=A+B(2 x+2)$
$x-3=A+2 B x+2 B$
On comparing,
$\left[\begin{array}{c} x=2 B x \Rightarrow B=\frac{1}{2} \\ -3=A+2 B \Rightarrow-3-2 \times \frac{1}{2}=A \Rightarrow A=-4 \end{array}\right]$
$=-4 \int \frac{d x}{x^{2}+2 x-4}+\frac{1}{2} \int \frac{\frac{d}{d x}\left(x^{2}+2 x-4\right)}{x^{2}+2 x-4}$ $\left[\begin{array}{l} x^{2}+2 x+1-4-1 \\ (x+1)^{2}-(\sqrt{5})^{2} \end{array}\right]$
$=-4 \int \frac{d x}{(x+1)^{2}-(\sqrt{5})^{2}}+\frac{1}{2} \log \left|x^{2}+2 x-4\right|$ $\left[\int \frac{d x}{x^{2}-a^{2}}=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+c\right]$
$=-\frac{2}{\sqrt{5}} \log \left|\frac{x+1-\sqrt{5}}{x+1+\sqrt{5}}\right|+\frac{1}{2} \log \left|x^{2}+2 x-4\right|+c$

Indefinite Integrals Exercise 18.19 Question 4

$I=\log \left|x^{2}+6 x+13\right|-\frac{9}{2} \tan ^{-1}\left(\frac{x+3}{2}\right)+c$
Hint: Using Formula $\int \frac{d x}{x^{2}-a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c \& \int \frac{d t}{t}=\log t+c$
Given: $\int \frac{2 x-3}{x^{2}+6 x+13} d x$
Solution: $\frac{d}{d x}\left(x^{2}+6 x+13\right)=2 x+6$
Let, $2 x-3=A(2 x+6)+B$
$2 x-3=2 A x+6 A+B$
On comparing,
$2 x=2 A x=>A=1$
$6 A+B=-3=>B=-9$
$I=\int \frac{2 x+6}{x^{2}+6 x+13} d x-9 \int \frac{d x}{x^{2}+6 x+13}$
$\text { Let, } x^{2}+6 x+13=t$
$(2 x+6) d x=d t$
$I=\int \frac{d t}{t}-9 \int \frac{d x}{x^{2}+2(3) x+3^{2}-3^{2}+13}$
$I=\int \frac{d t}{t}-9 \int \frac{d x}{(x+3)^{2}-9+13}$
$I=\int \frac{d t}{t}-9 \int \frac{d x}{(x+3)^{2}-4}$
$I=\int \frac{d t}{t}-9 \int \frac{d x}{(x+3)^{2}-2^{2}}$ $\left[\int \frac{d x}{x^{2}-a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c \& \int \frac{d t}{t}=\log t+c\right]$
$I=\log |t|-9 \times \frac{1}{2} \tan ^{-1}\left(\frac{x+3}{2}\right)+c$
$I=\log \left|x^{2}+6 x+13\right|-\frac{9}{2} \tan ^{-1}\left(\frac{x+3}{2}\right)+c$

Indefinite Integrals Exercise 18.19 Question 5

Answer: $-\frac{\sqrt{5}}{15} \tan ^{-1}\left|\frac{3 x-2}{\sqrt{5}}\right|+\frac{1}{2} \log \left|x^{2}-\frac{4}{3} x+1\right|+c$
Hint: Using Formula $\int \frac{d x}{x^{2}-a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c \& \int \frac{d t}{t}=\log t+c$
Given:$\int \frac{x-1}{3 x^{2}-4 x+3} d x$
Solution:
$\int \frac{x-1}{3 x^{2}-4 x+3} d x$
$\frac{1}{9} \int \frac{x-1}{x^{2}-\frac{4 x}{3}+1} d x$
$x-1=A+B \frac{d}{d x}\left(x^{2}-\frac{4 x}{3}+1\right)$
$x-1=A+B\left(2 x-\frac{4}{3}\right)$ $\left[\begin{array}{l} x=2 B x \Rightarrow B=\frac{1}{2} \\ -1=A-\frac{4}{3} B \Rightarrow-1+\frac{4}{3} \times \frac{1}{2}=A \\ -1+\frac{2}{3}=A \Rightarrow A=\frac{-1}{3} \end{array}\right]$
$I=\frac{-1}{3 \times 3} \int \frac{d x}{x^{2}-\frac{4}{3} x+1}+\frac{1}{3 \times 2} \int \frac{6 x-4}{x^{2}-\frac{4}{3} x+1} d x$
$I=\frac{-1}{9} \int \frac{d x}{x^{2}-\frac{4}{3} x+1}+\frac{1}{6} \int \frac{d t}{t}$ $\left[\begin{array}{c} \text { Let, } x^{2}-\frac{4}{3} x+1=t \\ \left(2 x-\frac{4}{3}\right) d x=d t \\ \frac{(6 x-4) d x}{3}=d t \end{array}\right]$
$I=\frac{-1}{9} \int \frac{d x}{x^{2}-\frac{4}{3} x+\frac{4}{9}+1-\frac{4}{9}}+\frac{1}{2} \int \frac{d t}{t}$
$I=-\frac{1}{9} \int \frac{d x}{\left(x-\frac{2}{3}\right)^{2}+\left(\frac{\sqrt{5}}{8}\right)^{2}}+\frac{1}{2} \int \frac{d t}{t}$ $\left[\int \frac{d x}{x^{2}-a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c \& \int \frac{d t}{t}=\log t+c\right]$
$I=\frac{-1}{9} \times \frac{1}{\frac{\sqrt{5}}{3}} \tan ^{-1}\left|\frac{x-\frac{2}{3}}{\frac{\sqrt{5}}{3}}\right|+\frac{1}{2} \log \left|x^{2}-\frac{4}{3} x+1\right|+c$
$I=\frac{-1}{3 \sqrt{5}} \tan ^{-1}\left|\frac{3 x-2}{\sqrt{5}}\right|+\frac{1}{2} \log \left|x^{2}-\frac{4}{3} x+1\right|+c$
$I=\frac{-3 \sqrt{5}}{3 \sqrt{5} \times 3 \sqrt{5}} \tan ^{-1}\left|\frac{3 x-2}{\sqrt{5}}\right|+\frac{1}{2} \log \left|x^{2}-\frac{4}{3} x+1\right|+c$
$I=\frac{-\sqrt{5}}{15} \tan ^{-1}\left|\frac{3 x-2}{\sqrt{5}}\right|+\frac{1}{2} \log \left|x^{2}-\frac{4}{3} x+1\right|+c$

Indefinite Integrals Exercise 18.19 Question 6

Answer: $\frac{-4}{3} \ln |x-2|-\frac{2}{3} \ln |x+1|+c$
Hint: You must know about how to solve integration
Given: $\int \frac{2 x}{2+x-x^{2}} d x$
Solution: $\int \frac{2 x}{2+x-x^{2}} d x$
$=-\int \frac{2 x}{x^{2}-x-2} d x$
$=-\int \frac{2 x}{(x-2)(x+1)} d x$
$\frac{A}{(x-2)}+\frac{B}{(x+1)}=\frac{2 x}{(x-2)(x+1)}$
$A(x+1)+B(x+2)=2 x$
$A x+A+B x+2 B=2 x$
On comparing,

$\left(\begin{array}{c|c} A x+B x=2 x & A-2 B=0 \\ A+B=2 & A=2 B \end{array}\right)$
$2 B+B=2$
$B=\frac{2}{3}, A=\frac{4}{3}$
$\int-\frac{4}{3}\left(\frac{1}{x-2}\right) d x-\frac{2}{3} \int \frac{1}{x+1} d x$
$=-\frac{4}{3} \log |x-2|-\frac{2}{3} \log |x+1|+c$

Indefinite Integrals Exercise 18.19 Question 7

Answer:$-\frac{1}{2} \ln \left(3 x^{2}+4 x+2\right)+\frac{3}{\sqrt{2}} \tan ^{-1}\left(\frac{3 x+2}{\sqrt{2}}\right)+c$
Hint:Use $\int \frac{P n+q}{a x^{2}+b x+c}$
Given: $\int \frac{1-3 x}{3 x^{2}+4 x+2} d x$
Solution:
$I=\int \frac{1-3 x}{3 x^{2}+4 x+2} d n$
$\int \frac{P n+q}{a x^{2}+b x+c} d x$
$\text { Numerator }=A\left(\frac{d}{d x}(\text { deno min ator })\right)+B$
$1-3 x=A\left(\frac{d}{d x}\left(3 x^{2}+4 x+2\right)\right)+B$
$1-3 x=A(6 x+4)+B$
$1-3 x=6 A x+4 A+B$
On comparing,
$6 A=-3 \Rightarrow A=-\frac{3}{6} \Rightarrow A=\frac{-1}{2}$
$4 A+B=1 \Rightarrow 4\left(\frac{-1}{2}\right)+B=1 \Rightarrow-2+B=1 \Rightarrow B=3$
$I=\int \frac{\left(-\frac{1}{2}\right)(6 x+4)+3}{3 x^{2}+4 x+2} d x$
$I=\int-\frac{1}{2} \frac{(6 x+4)}{3 x^{2}+4 x+2} d x+\int \frac{3}{3 x^{2}+4 x+2} d x$
$=\frac{-1}{2} \int \frac{(6 x+4)}{3 x^{2}+4 x+2} d x+\int \frac{3}{3\left(x^{2}+\frac{4}{3} x+\frac{2}{3}\right)} d x$
$=\frac{-1}{2} \int \frac{(6 x+4)}{3 x^{2}+4 x+2} d x+\int \frac{1}{x^{2}+\frac{4}{3} x+\frac{4}{9}-\frac{4}{9}+\frac{2}{3}} d x$
$=\frac{-1}{2} \int \frac{\frac{d}{d x}\left(3 x^{2}+4 x+2\right)}{3 x^{2}+4 x+2} d x+\int \frac{1}{\left(x+\frac{2}{3}\right)^{2}+\left(\frac{\sqrt{2}}{3}\right)^{2}} d x$
\begin{aligned} &3 x^{2}+4 x+2=t \\ &\frac{d\left(3 x^{2}+4 x+2\right)}{d x}=d t \end{aligned}
$=-\frac{1}{2} \int \frac{d t}{t}+\frac{1}{\frac{\sqrt{2}}{3}}\left[\frac{\tan ^{-1} x+\frac{2}{3}}{\frac{\sqrt{2}}{3}}\right]\left[\int \frac{1}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{2}+c\right]$
$=-\frac{1}{2} \ln \left(3 x^{2}+4 x+2\right)+\frac{3}{\sqrt{2}} \tan ^{-1}\left(\frac{3 x+2}{\sqrt{2}}\right)+c$

Indefinite Integrals Exercise 18.19 Question 8

Answer: $\log \left|x^{2}-x-2\right|+2 \log \left|\frac{x-2}{x+1}\right|+c$
Hint: You must know about how to solve integration
Given:$\int \frac{2 x+5}{x^{2}-x-2} d x$
Solution:
Let $I=\int \frac{2 x+5}{x^{2}-x-2} d x$
$=\int \frac{2 x-1+6}{x^{2}-x-2} d x$
$=\int \frac{2 x-1}{x^{2}-x-2} d x+\int \frac{6}{x^{2}-x-2} d x$
$I_{1}=\int \frac{2 x-1}{x^{2}-x-2} d x$
Let
$x^{2}-x-2=t$
$\Rightarrow(2 x-1) d x=d t$
\begin{aligned} &I_{1}=\int \frac{d t}{t}=\log |t|+c_{1} \\ &=\log \left|x^{2}-x-2\right|+c_{1} \end{aligned}
\begin{aligned} &I_{2}=\int \frac{6}{x^{2}-x-2} d x \\ &=6 \int \frac{d x}{x^{2}-x+\frac{1}{4}-\frac{1}{4}-2} \\ &=6 \int \frac{d x}{\left(x-\frac{1}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}}\left[\int \frac{d x}{x^{2}-a^{2}}=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+c\right] \\ &=6 \times \frac{1}{3} \log \left|\frac{x-\frac{1}{2}-\frac{3}{2}}{x-\frac{1}{2}+\frac{3}{2}}\right|+c_{2} \\ &=2 \log \left|\frac{2 x-4}{2 x+2}\right|+c_{2} \end{aligned}
So,$I=I_{1}+I_{2}$
$=\log \left|x^{2}-x-2\right|+2 \log \left|\frac{x-2}{x+1}\right|+c$

Indefinite Integrals Exercise 18.19 Question 9

Answer:$\frac{a}{4} \ln \left|x^{4}+c^{2}\right|+\frac{b}{2 c} \tan ^{-1} \frac{x^{2}}{c}+f$
Hint : Find $I_{1}\: and \: I_{2}$
Given: $\int \frac{a x^{3}+b x}{x^{4}+c^{2}} d x$
Solution: $\int \frac{a x^{3}+b x}{x^{4}+c^{2}} d x$
\begin{aligned} I &=\int \frac{a x^{3}}{x^{4}+c^{2}} d x+\int \frac{b x}{x^{4}+c^{2}} d x \\ I_{1} &=\int \frac{a x^{3}}{x^{4}+c^{2}} d x \end{aligned}
Let,$x^{4}+c^{2}=t$
\begin{aligned} &4 x^{3} d x=d t \\ &x^{3} d x=\frac{1}{4} d t \end{aligned}
$\Rightarrow I_{1}=\frac{a}{4} \int \frac{d t}{t}=\frac{a}{4} \ln t+C_{1}$
\begin{aligned} &I_{2}=\int \frac{b x}{x^{4}+c^{2}} d x \\ &\text { Let } x^{2}=v \\ &2 x d x=d v \end{aligned}
$I_{2}=\frac{b}{2} \int \frac{d v}{v^{2}+c^{2}}$
$=\frac{b}{2} \tan ^{-1}\left(\frac{v}{c}\right) \times \frac{1}{c}+c_{2} \quad\left[\int \frac{1}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{2}+c\right]$
$I_{2}=\frac{b}{2 c} \tan ^{-1}\left(\frac{x^{2}}{c}\right)+c_{2}$
\begin{aligned} &I=I_{1}+I_{2} \\ &I=\frac{a}{4} \ln \left|x^{4}+c^{2}\right|+c_{1}+\frac{b}{2 c} \tan ^{-1}\left(\frac{x^{2}}{c}\right)+c_{2} \\ &c_{1}+c_{2}=f \\ &I=\frac{a}{4} \ln \left|x^{4}+c^{2}\right|+\frac{b}{2 c} \tan ^{-1}\left(\frac{x^{2}}{c}\right)+f \end{aligned}

Indefinite Integrals Exercise 18.19 Question 10

Answer:$\frac{4}{2-\sin x}+3 \log |2-\sin x|+c$
Hint: Use integration method
Given: $\int \frac{(3 \sin x-2) \cos x}{5-\cos ^{2} x-4 \sin x} d x$
Solution: $I=\int \frac{(3 \sin x-2) \cos x}{5-\cos ^{2} x-4 \sin x} d x$
$=\int \frac{(3 \sin x-2) \cos x}{5-\left(1-\sin ^{2} x\right)-4 \sin x} d x$
$=\int \frac{(3 \sin x-2) \cos x}{4+\sin ^{2} x-4 \sin x} d x$
$=\int \frac{(3 \sin x-2) \cos x}{(2-\sin x)^{2}} d x$
\begin{aligned} &\text { Let } t=2-\sin x \\ &\sin x=2-t \\ &\qquad d t=-\cos x d x \end{aligned}
\begin{aligned} &I=\int \frac{3(2-t)-2}{t^{2}}(-d t) \\ &=-\int \frac{4-3 t}{t^{2}} d t \end{aligned}
\begin{aligned} &=-4 \int \frac{d t}{t^{2}}+3 \int \frac{1}{t} d t \\ &I=-4\left(\frac{-1}{t}\right)+3 \log t+c_{1} \\ &I=\frac{4}{2-\sin x}+3 \log |2-\sin x|+c \end{aligned}

Indefinite Integrals Exercise 18 .19 Question 11

Answer:$\frac{1}{4} \ln \left(2 x^{2}+6 x+5\right)+\frac{1}{2} \tan ^{-1}(2 x+3)+c$
Hint: Multiply and Divide by 4
Given: $\int \frac{x+2}{2 x^{2}+6 x+5} d x$
Solution:Multiply and divide by 4
$\frac{1}{4} \int \frac{4 x+6}{2 x^{2}+6 x+5} d x+\frac{2}{4} \int \frac{d x}{2 x^{2}+6 x+5}$ .........(1)
\begin{aligned} &I_{1}=\int \frac{4 x+6}{2 x^{2}+6 x+5} d x \\ &2 x^{2}+6 x+5=t \\ &(4 x+6) d x=d t \\ &I_{1}=\int \frac{d t}{t}=\ln t+c=\ln \left(2 x^{2}+6 x+5\right)+c_{1} \end{aligned}
\begin{aligned} &I_{2}=\frac{1}{2} \int \frac{d x}{2 x^{2}+6 x+5} \\ &I_{2}=\frac{1}{2} \int \frac{d x}{x^{2}+3 x+\frac{9}{4}+\frac{5}{2}-\frac{9}{4}} \end{aligned}
$\frac{1}{2} \int \frac{d x}{\left(x+\frac{3}{2}\right)^{2}+\frac{1}{4}}$
\begin{aligned} &\left(x+\frac{3}{2}\right)=u \Rightarrow 1 d x=d u \\ &I_{2}=\frac{1}{2} \int \frac{d u}{u^{2}+\left(\frac{1}{2}\right)^{2}} \\ &\frac{1}{2}\left[\frac{1}{\frac{1}{2}} \tan ^{-1}\left(\frac{u}{\frac{1}{2}}\right)\right]+c_{2}\left[\int \frac{1}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{2}\right. \\ &\begin{array}{l} \tan ^{-1}\left[2 \tan ^{-1}2\left(x+\frac{3}{2}\right)\right]+c_{2} \\ I_{2}=\tan ^{-1}(2 x+3)+c_{2} \\ I=\frac{1}{4} \ln \left(2 x^{2}+6 x+5\right)+c_{1}+\frac{1}{2} \tan ^{-1}(2 x+3)+c_{2} \\ I=\frac{1}{4} \ln \left(2 x^{2}+6 x+5\right)+\frac{1}{2} \tan ^{-1}(2 x+3)+c \end{array} \end{aligned}

Indefinite Integrals Exercise 18.19 Question 12

Answer: $\frac{5}{6} \ln \left|3 x^{2}+2 x+1\right|-\frac{11}{3 \sqrt{2}} \tan ^{-1} \frac{3 x+1}{\sqrt{2}}+c$
Hint: Find value of M and N
Given: $\int \frac{5 x-2}{1+2 x+3 x^{2}} d x$
Solution: $I=\int \frac{5 x-2}{1+2 x+3 x^{2}} d x$
\begin{aligned} &5 x-2=M \frac{d}{d x}\left(1+2 x+3 x^{2}\right)+N \\ &5 x-2=M(2+6 x)+N \\ &5 x-2=2 M+6 M x+N \end{aligned}
On comparing,
\begin{aligned} &6 M=5 \Rightarrow M=\frac{5}{6} \\ &2 M+N=-2 \Rightarrow 2\left(\frac{5}{6}\right)+N=-2 \\ &\Rightarrow \frac{5}{3}+N=-2 \Rightarrow N=-2-\frac{5}{3}=-\frac{11}{3} \end{aligned}
$I=\frac{5}{6} \int \frac{2+6 x}{1+2 x+3 x^{2}} d x-\frac{11}{3} \int \frac{d x}{1+2 x+3 x^{2}}$ ................................(1)
\begin{aligned} &I_{1}=\int \frac{2+6 x}{1+2 x+3 x^{2}} d x \\ &\text { let, } \quad t=1+2 x+3 x^{2} \\ &\Rightarrow 2+6 x=\frac{d t}{d x} \\ &\Rightarrow(2+6 x) d x=d t \end{aligned}
$d x=\frac{d t}{2+6 x}$
\begin{aligned} &I_{1}=\int \frac{d t}{t}=\log |t|+c \\ &I_{2}=\int \frac{d x}{1+2 x+3 x^{2}} \end{aligned}
\begin{aligned} &=3\left(x^{2}+2 \cdot x \cdot \frac{1}{3}+\left(\frac{1}{3}\right)^{2}-\left(\frac{1}{3}\right)^{2}+\frac{1}{3}\right) \\ &=3\left(\left(x+\frac{1}{3}\right)^{2}+\frac{1}{3}-\frac{1}{9}\right) \\ &=3\left(\left(x+\frac{1}{3}\right)^{2}+\frac{3-1}{9}\right) \\ &=3\left(\left(x+\frac{1}{3}\right)^{2}+\left(\frac{\sqrt{2}}{3}\right)^{2}\right) \end{aligned}
From eq 1
$I=\frac{5}{6} \int \frac{d t}{t}-\frac{11}{3} \int \frac{d x}{3\left(x+\frac{1}{3}\right)^{2}+\left(\frac{\sqrt{2}}{3}\right)^{2}}$
$=\frac{5}{6} \log |t|-\frac{11}{9} \int \frac{d x}{\left(x+\frac{1}{3}\right)^{2}+\left(\frac{\sqrt{2}}{3}\right)^{2}}$ $\int \frac{d x}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c$
$=\frac{5}{6} \log \left|3 x^{2}+2 x+1\right|-\frac{11}{9} \times \frac{1}{\frac{\sqrt{2}}{3}} \tan ^{-1} \frac{\left(x+\frac{1}{3}\right)}{\frac{\sqrt{2}}{3}}$
$I=\frac{5}{6} \log \left|3 x^{2}+2 x+1\right|-\frac{11}{2 \sqrt{3}} \tan ^{-1} \frac{(3 x+1)}{\sqrt{2}}+c$

Indefinite Integrals Exercise 18.19 Question 13

$\frac{1}{3} \log |3 x-2|+c$
Hint: Solve integration by equation
Given:
\begin{aligned} &\int \frac{x+5}{3 x^{2}+13 x-10} d x \\ &=\int \frac{x+5}{3 x^{2}+15 x-2 x-10} d x \\ &=\int \frac{x+5}{3 x(x+5)-2(x+5)} d x \end{aligned}
\begin{aligned} &=\int \frac{x+5}{(3 x-2)(x+5)} d x \\ &=\int \frac{d x}{3 x-2} \end{aligned} .........(1)
\begin{aligned} &\text { Let, } 3 x-2=t \\ &\qquad \begin{array}{r} 3 d x=d t \\ d x=\frac{d t}{3} \end{array} \end{aligned}
\begin{aligned} &=\int \frac{d t}{3 t}+c \\ &=\frac{1}{3} \log |3 x-2|+c \end{aligned}

Indefinite Integrals Exercise 18.19 Question 14

Answer: $10 \ln |\sin x-4|-7 \ln |\sin x-3|+c$
Hint : Let $\sin x=t$
Given: $\int \frac{(3 \sin x-2) \cos x}{13-\cos ^{2} x-7 \sin x} d x$
Solution: $\text { Let } \sin x=t$
$\cos x d x=d t$
$\int \frac{(3 \sin x-2) \cos x}{13-\left(1-\sin ^{2} x\right)-7 \sin x} d x$ $\left(\cos ^{2} x=1-\sin ^{2} x\right)$
$\int \frac{(3 t-2) d t}{13-1+t^{2}-7 t}=\int \frac{(3 t-2) d t}{t^{2}-7 t+12}$ $\left[\begin{array}{l} t^{2}-7 t+12=(t-4)(t-3) \\ a x^{2}+b x+c, x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \end{array}\right]$
$\int \frac{3 t-2}{(t-4)(t-3)} d t=\frac{A}{t-4}+\frac{B}{t-3}$ ..........(1)
\begin{aligned} &3 t-2=\frac{A}{t-4}+\frac{B}{t-3} \\ &3 t-2=A(t-3)+B(t-4) \\ &3 t-2=A t-3 A+B t-4 B \end{aligned}
On comparing,
$3 \mathrm{t}=\mathrm{At}+\mathrm{Bt} \Rightarrow \mathrm{A}+\mathrm{B}=3 \Rightarrow>\mathrm{A}=3-\mathrm{B}$
$-2=-3 A-4 B=23 A+4 B=2=>3(3-B)+4 B=2=>B=-7$
On solving both
$A=3-(-7) \Rightarrow A=10$
In eqn (1)
\begin{aligned} & \int\left[\frac{10}{t-4}+\frac{-7}{t-3}\right] d t \\ =& \int \frac{10}{t-4} d t-\int \frac{7}{t-3} d t \end{aligned} $\int \frac{1}{t} d t=\ln |t|+c$
\begin{aligned} &10 \ln |t-4|-7 l n|t-3|+c \\ &=10 \ln |t-4|-7 \ln |t-3|+c \\ &=10 \ln |\sin x-4|-7 \ln |\sin x-3|+c \end{aligned}

Indefinite Integrals Exercise 18.19 Question 15

Answer:$\log \frac{|3 x+4|}{3}+c$
Hint: Solve Integration
Given: $\int \frac{x+7}{3 x^{2}+25 x+28} d x$
Solution:
\begin{aligned} &\int \frac{x+7}{3 x^{2}+25 x+28} d x \\ &=\int \frac{x+7}{3 x^{2}+21 x+4 x+28} d x \\ &=\int \frac{x+7}{3 x(x+7)+4(x+7)} d x \\ &=\int \frac{x+7}{(3 x+4)(x+7)} d x \\ &=\int \frac{d x}{3 x+4} \end{aligned}
Let,\begin{aligned} &3 x+4=t \\ &3 d x=d t \end{aligned}
$d x=\frac{d t}{3}$
\begin{aligned} &=\int \frac{d t}{3 t}+c \\ &=\frac{1}{3} \log |3 x+4|+c \end{aligned}

Indefinite Integrals Exercise 18.19 Question 16

Answer: $\frac{14}{9} \log |x-3|+\frac{13}{9} \ln |x+6|+c$
Hint: You know about how to solve integration
Given:$\int \frac{3 x+5}{x^{2}+3 x-18} d x$
Explanation:
We have, $\int \frac{3 x+5}{x^{2}+3 x-18} d x$
\begin{aligned} &\Rightarrow \int \frac{3 x+5}{x^{2}+6 x-3 x-18} d x \\ &\Rightarrow \int \frac{3 x+5}{x(x+6)-3(x+6)} d x \\ &\Rightarrow \int \frac{3 x+5}{(x-3)(x+6)} d x \end{aligned}
Now, Using Partial fraction,
\begin{aligned} &\frac{3 x+5}{(x-3)(x+6)}=\frac{A}{(x+6)}+\frac{B}{(x-3)} \\ &\frac{3 x+5}{(x-3)(x+6)}=\frac{A(x-3)+B(x+6)}{(x-3)(x+6)} \\ &\frac{3 x+5}{(x-3)(x+6)}=\frac{A x-3 A+B x+6 B}{(x-3)(x+6)} \\ &\Rightarrow \frac{3 x+5}{(x-3)(x+6)}=\frac{(A+B) x-3 A+6 B}{(x-3)(x+6)} \end{aligned}
On comparing, we got
$A+B=3$ ......(1)
and
\begin{aligned} &-3 A+6 B=5 \\ &(A+B=3) \times-3 \end{aligned} ........(2)
We get,
$-3 A-3 B=-9 \ldots .(3)$
Subtract (2) from (3)
\begin{aligned} &(-3 A-3 B)-(-3 A+6 B)=(-9)-(5) \\ &-3 A-3 B+3 A-6 B=-9-5 \\ &-9 B=-14 \\ &B=\frac{14}{9} \end{aligned}
Put value of B in (1)
\begin{aligned} &A+B=3 \\ &A+\frac{14}{9}=3 \\ &A=3-\frac{14}{9} \Rightarrow \frac{27-14}{9} \\ &A=\frac{13}{9} \& B=\frac{14}{9} \end{aligned}
Substituting value of A and B in
\begin{aligned} &\frac{3 x+5}{(x-3)(x+6)}=\frac{A}{(x+6)}+\frac{B}{(x-3)} \\ &\frac{3 x+5}{(x-3)(x+6)}=\frac{13}{9} \frac{1}{(x+6)}+\frac{14}{9} \frac{B}{(x-3)} \end{aligned}
Integrating both sides
$\int \frac{3 x+5}{(x-3)(x+6)} d x=\frac{13}{9} \int \frac{1}{(x+6)} d x+\frac{14}{9} \int \frac{1}{(x-3)} d x$ $\left[\int \frac{1}{x+a} d x=\log |x+a|\right]$
$\int \frac{3 x+5}{(x-3)(x+6)} d x=\frac{13}{9} \log |x+6|+\frac{14}{9} \log |x-3|+c$

Indefinite Integrals Exercise 18 .19 Question 17

Answer: $\frac{1}{4} \ln \left|x^{4}+x^{2}+1\right|-\frac{1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 x^{2}+1}{\sqrt{3}}\right)+c$
Hint: You know about the integration of $\int x^{3}$
Given: $\int \frac{x^{3}}{x^{4}+x^{2}+1} d x$
Solution: $\int \frac{x^{3}}{x^{4}+x^{2}+1} d x$
\begin{aligned} &\int \frac{x^{3}}{x^{4}+x^{2}+1} d x \\ &=\frac{1}{2} \int \frac{2 x^{3}}{x^{4}+x^{2}+1} d x \\ &=\frac{1}{2} \int \frac{x^{2}}{x^{4}+x^{2}+1} 2 x d x \end{aligned}
Let $x^{2}=t$
$2 x d x=d t$
\begin{aligned} &\frac{1}{2} \int \frac{t}{t^{2}+t+1} d t \\ &\Rightarrow \frac{1}{2} \int \frac{2 t+1-1}{2\left(t^{2}+t+1\right)} d t \\ &\Rightarrow \frac{1}{4} \int \frac{2 t+1}{t^{2}+t+1} d t-\int \frac{d t}{t^{2}+t+1} \\ &\Rightarrow \frac{1}{4} \int \frac{2 t+1}{t^{2}+t+1} d t-\int \frac{d t}{t^{2}+t+\frac{1}{4}-\frac{1}{4}} \end{aligned}
$\Rightarrow \frac{1}{4} \int \frac{2 t+1}{t^{2}+t+1} d t-\int \frac{d t}{\left(\frac{3}{4}\right)+\left(t+\frac{1}{2}\right)^{2}}$
Let, \begin{aligned} t^{2}+2 t+1 &=u \\ 2 t+1 d t &=d u \end{aligned}
\begin{aligned} &=\frac{1}{4} \int \frac{d u}{u}-\int \frac{d t}{\left(\frac{\sqrt{3}}{2}\right)^{2}+\left(t+\frac{1}{2}\right)^{2}} \\ &=\frac{1}{4}\left[\ln u-\frac{1}{\frac{\sqrt{3}}{2}} \tan ^{-1}\left(\frac{t+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)\right]+c \end{aligned}
\begin{aligned} &=\frac{1}{4}\left(\ln \left(t^{2}+t+1\right)-\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 t+1}{\sqrt{3}}\right)\right)+c \\ &=\frac{1}{4} \ln \left|x^{4}+x^{2}+1\right|-\frac{1}{4} \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x^{2}+1}{\sqrt{3}}\right)+c \\ &=\frac{1}{4} \ln \left|x^{4}+x^{2}+1\right|-\frac{1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 x^{2}+1}{\sqrt{3}}\right)+c \end{aligned}

Indefinite Integrals Exercise 18.19 Question 18

Answer:$\frac{-\sqrt{5}}{5} \log \left|\frac{x^{2}+1-\sqrt{5}}{x^{2}+1+\sqrt{5}}\right|+\frac{1}{4} \log \left|x^{4}-2 x^{2}-4\right|+c$
Hint : Let $x^{2}=t$
Given: $\int \frac{x^{3}-3 x}{x^{4}+2 x^{2}-4} d x$
Solution: $\int \frac{x\left(x^{2}-3\right)}{\left(x^{2}\right)^{2}+2 x^{2}-4} d x$
Let $x^{2}=t$
$x d x=\frac{d t}{2}$
$\frac{1}{2} \int \frac{(t-3)}{t^{2}+2 t-4} d t$
Multiply and divide by 2
\begin{aligned} &=\frac{1}{4} \int \frac{2 t-6}{t^{2}+2 t-4} d t \\ &=\frac{1}{4} \int \frac{2 t+2-8}{t^{2}+2 t-4} d t \\ &=\frac{1}{4} \int \frac{2 t+2}{t^{2}+2 t-4} d t-\frac{1}{4} \int \frac{8}{t^{2}+2 t-4} d t \\ &=\frac{1}{4} \int \frac{2 t+2}{t^{2}+2 t-4} d t-2 \int \frac{1}{t^{2}+2 t-4} d t \\ &=\frac{1}{4} \int \frac{2 t+2}{t^{2}+2 t-4} d t-2 \int \frac{1}{t^{2}+2 t+1-5} d t \\ &=\frac{1}{4} \int \frac{2 t+2}{t^{2}+2 t-4} d t-2 \int \frac{1}{(t+1)^{2}-\sqrt{5}^{2}} d t \end{aligned}
$I=\frac{1}{4}\log \left | t^{2}+2t-4 \right |-2\times \frac{1}{2\sqrt{5}}\log (\frac{t+1-\sqrt{5}}{t+1+\sqrt{5}})+c$
$I=\frac{1}{4}\log \left | x^{4}+2x^{2}-4 \right |-\frac{1}{\sqrt{5}}\log (\frac{x^{2}+1-\sqrt{5}}{x^{2}+1+\sqrt{5}})+c$

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1. How to ace the chapter on indefinite integrals?

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2. Indefinite Integral

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