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Define Indefinite Integrals in mathematics?

An indefinite integral refers to any integral without an upper and lower limit.

What are the different types of Integrals?

The integrals are two types and these are: Definite Integral Indefinite Integral

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

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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

Updated on 24 Jan 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.

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RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise
Indefinite Integrals Excercise:18.19
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RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

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

Answer:
$I=\log \left|x^{2}+6 x+13\right|-\frac{9}{2} \tan ^{-1}\left(\frac{x+3}{2}\right)+c$
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

Answer:
$\frac{1}{3} \log |3 x-2|+c$
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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