RD Sharma Class 12 Exercise 18.32 Indefinite Integrals Solutions Maths - Download PDF Free Online

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RD Sharma Class 12 Exercise 18.32 Indefinite Integrals Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 18.32 Indefinite Integrals Solutions Maths - Download PDF Free Online

Kuldeep MauryaUpdated on 24 Jan 2022, 10:54 AM IST

RD Sharma books are considered as some of the best materials to study for exams. Majority of CBSE schools and teachers follow this book for their lectures as well as the set of question papers. This is why it is a leading choice among students as it is informative and detailed. RD Sharma Solutions Indefinite Integrals Ex 18.32 material is prepared by a group of experts to help students get a better understanding of the subject. The Class 12 RD Sharma chapter 18 exercise 18.32 solution meets a fantastic essential of maths questions. It contains step-by-step solutions for each answer making it easier for students to understand.

RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.32

Indefinite Integrals Excercise 18.32 Question 2

Answer : - $\frac{1}{\sqrt{5}} \log \left|\frac{\sqrt{2 x+3}-\sqrt{5}}{\sqrt{2 x+3}+\sqrt{5}}\right|+C$
Hint :- Use substitution method to solve this integral.
Given :- $\int \frac{1}{(x-1) \sqrt{2 x+3}} d x$
Sol : - Let $\mathrm{I}=\int \frac{1}{(x-1) \sqrt{2 x+3}} d x$
Put $2 x+3=t^{2} \Rightarrow 2 d x=2 t d t \rightarrow d x=t dt$ .then,
$\begin{aligned} &=\int \frac{1}{\left(\frac{t^{2}-5}{2}\right)} d t \\ &=\int \frac{2}{t^{2}-5} d t \\ &=\int \frac{2}{t^{2}-\left(\sqrt{\left.5^{2}\right)}\right.} d t \end{aligned}$
$\begin{aligned} &=2 \cdot \frac{1}{2 \sqrt{5}} \log \left|\frac{t-\sqrt{5}}{t+\sqrt{5}}\right|+C \quad\left(\because \int \frac{1}{x^{2}-a^{2}} d x=1 / 2 a \log \left|\frac{x-a}{x+a}\right|+C\right) \\ &=\frac{1}{\sqrt{5}} \log \left|\frac{\sqrt{2 x+3}-\sqrt{5}}{\sqrt{2 x+3}+\sqrt{5}}\right|+C \quad(\because t=\sqrt{2 x+3}) \text { Ans.. } \end{aligned}$ .

Indefinite Integrals Excercise 18.32 Question 3

Answer : - $2 \sqrt{x+2}+\frac{2}{\sqrt{3}} \log \left|\frac{\sqrt{x+2}-\sqrt{3}}{\sqrt{x+2}+\sqrt{3}}\right|+C$
Hint :- Use substitution method to solve this integral.
Given :- $\int \frac{x+1}{(x-1) \sqrt{x+2}} d x \\$
Sol : -
$\begin{aligned} &\text { Let } \mathrm{I}=\int \frac{x+1}{(x-1) \sqrt{x+2}} d x \\ &=\int \frac{x-1+2}{(x-1) \sqrt{x+2}} d x \\ &\Rightarrow \mathrm{I}=\int \frac{(x-1)+2}{(x-1) \sqrt{x+2}} d x \\ &\Rightarrow \mathrm{I}=\int\left\{\frac{(x-1)}{(x-1) \sqrt{x+2}}+\frac{2}{(x-1) \sqrt{x+2}}\right\} d x \end{aligned}$
$\Rightarrow \mathrm{I}=\int \frac{1}{\sqrt{x+2}} d x+\int \frac{2}{(x-1) \sqrt{x+2}} d x$
Now, I = I1 +I2 ........................ (I)
Where $\mathrm{I}_{1}=\int \frac{1}{\sqrt{x+2}} d x \text { and } \mathrm{I}_{2}=\int \frac{2}{(x-1) \sqrt{x+2}}dx$
Therefore, $I_{1}=\int \frac{1}{\sqrt{x+2}} d x$
Let $\mathrm{x}+2=\mathrm{u} \Rightarrow \mathrm{d} \mathrm{x}=\mathrm{du}$ Then,
$\begin{aligned} \mathrm{I}_{1} &=\int \frac{1}{u^{\frac{1}{2}}} d u=\int u^{\frac{-1}{2}} \mathrm{du}=\frac{u^{\frac{-1}{2}}+1}{\frac{-1}{2}+1}+C \quad\left(\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right) \\ &=\frac{u^{\frac{1}{2}}}{\frac{1}{2}}+\mathrm{C}_{1} \\ &=2 \sqrt{x+2}+\mathrm{C}_{1} \ldots \ldots \ldots \ldots \text { (II) } \end{aligned}$
And $I_{2}=\int \frac{2}{(x-1) \sqrt{x+2}} d x$
Let $x+2=t^{2}=>d x=2 t d t$. Then,
$\begin{aligned} \mathrm{I}_{2} &=\int \frac{2}{\left(t^{2}-2-1\right) \sqrt{t^{2}}} 2 \mathrm{t} \mathrm{d} \mathrm{t}=4 \int \frac{t d t}{\left(t^{2}-3\right) t} & &\left(\because \mathrm{t}^{2}-2=\mathrm{x}\right) \\ &=4 \int \frac{1}{\left(t^{2}\right)-\sqrt{3^{2}}} \mathrm{dt}=\frac{4}{2 \cdot \sqrt{3}} \log \left|\frac{t-\sqrt{3}}{t+\sqrt{3}}\right|+\mathrm{C} & &\left(\because \int \frac{1}{x^{2}-a^{2}} \mathrm{~d} \mathrm{x}=1 / 2 \mathrm{a} \log \left|\frac{x-a}{x+a}\right|+\mathrm{C}\right) \\ &=\frac{2}{\sqrt{3}} \log \left|\frac{\sqrt{x+2}-\sqrt{3}}{\sqrt{x+2}+\sqrt{3}}\right|+\mathrm{C}_{2} \quad &(\because \mathrm{t}=\sqrt{x+2}) \ldots \ldots \ldots . \mathrm{III}) \end{aligned}$

Put the value of Equ.(II) and (III) in (I) then,
$\begin{aligned} &\mathrm{I}=2 \sqrt{x+2}+\mathrm{C}_{1}+\frac{2}{\sqrt{3}} \log \left|\frac{\sqrt{x+2}-\sqrt{3}}{\sqrt{x+2}+\sqrt{3}}\right|+\mathrm{C}_{2} \\ &\quad=2 \sqrt{x+2}++\frac{2}{\sqrt{3}} \log \left|\frac{\sqrt{x+2}-\sqrt{3}}{\sqrt{x+2}+\sqrt{3}}\right|+C \quad\left(\because \mathrm{C}=\mathrm{C}_{1}+\mathrm{C}_{2}\right) \text { Ans.. } \end{aligned}$

Indefinite Integrals Excercise 18.32 Question 4

Answer :$\frac{2}{3}(x+2)^{\frac{3}{2}}-2 \sqrt{x+2}+\frac{1}{\sqrt{3}} \log \left|\frac{\sqrt{x+2}-\sqrt{3}}{\sqrt{x+2}+\sqrt{3}}\right|+C$
Hint :- Use substitution method and special integration formula to solve this problem.
Given :- $\int \frac{x^{2}}{(x-1) \sqrt{x+2}} d x$
Sol :-
$\begin{aligned} \text { let } I &=\int \frac{x^{2}}{(x-1) \sqrt{x+2}} d x \\ &=\int \frac{\left(x^{2}-1\right)+1}{(x-1) \sqrt{x+2}} d x=\int \frac{\left(x^{2}-1\right)+1}{(x-1) \sqrt{x+2}} d x \\ &=\int\left\{\frac{x^{2}-1}{(x-1) \sqrt{x+2}}+\frac{1}{(x-1) \sqrt{x+2}}\right\} d x \\ &=\int\left\{\frac{(x-1)(x+1)}{(x-1) \sqrt{x+2}}+\frac{1}{(x-1) \sqrt{x+2}}\right\} d x \quad\quad\quad\quad\left(\because a^{2}-b^{2}=(a+b)(a-b)\right) \end{aligned}$
$\begin{aligned} &=\int\left\{\frac{x+1}{\sqrt{x+2}}+\frac{1}{(x-1) \sqrt{x+2}}\right\} \mathrm{d} \mathrm{x} \\ &=\int\left\{\frac{x+2-1}{\sqrt{x+2}}+\frac{1}{(x-1) \sqrt{x+2}}\right\} \mathrm{d} \mathrm{x} \\ &\quad=\int\left\{\frac{(x+2)-1}{\sqrt{x+2}}+\frac{1}{(x-1) \sqrt{x+2}}\right\} \mathrm{d} \mathrm{x} \\ &\quad=\int\left\{\frac{(x+2)}{\sqrt{x+2}}-\frac{1}{\sqrt{x+2}}+\frac{1}{(x-1) \sqrt{x+2}}\right\} \mathrm{d} \mathrm{x} \end{aligned}$
$\begin{aligned} &=\int\left\{(x+2)^{1-\frac{1}{2}}-\frac{1}{\sqrt{x+2}}+\frac{1}{(x-1) \sqrt{x+2}}\right\} \mathrm{d} \mathrm{x} \\ &=\int\left\{\sqrt{x+2}-\frac{1}{\sqrt{x+2}}+\frac{1}{(x-1) \sqrt{x+2}}\right\} \mathrm{d} \mathrm{x} \\ &\mathrm{I}=\int \sqrt{x+2} d x-\int \frac{1}{\sqrt{x+2}} d x+\int \frac{1}{(x-1) \sqrt{x+2}} \mathrm{~d} \mathrm{x} \ldots \ldots \text { (I) } \end{aligned}$
Now,$\int \sqrt{x+2}dx$
Put $\mathrm{x}+2=\mathrm{p}=>\mathrm{d} \mathrm{x}=\mathrm{d} \mathrm{p}$ then
$\begin{aligned} \int \sqrt{x+2} d x &=\int \sqrt{p} d p=\int p^{\frac{1}{2} d}=\frac{p \frac{1}{2}+1}{\frac{1}{2}+1}+\mathrm{C}_{1} & &\left(\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right) \\ &=\frac{p^{\frac{8}{2}}}{3 / 2}=2 / 3(x+2)^{3 / 2}+\mathrm{C}_{1} \ldots \ldots . . .\left(2 \right ) &(\because \mathrm{p}=\mathrm{x}+2) \end{aligned}$
And, $\int \frac{1}{\sqrt{x+2}} d x$
Put $\mathrm{x}+2=\mathrm{U}=>\mathrm{d} \mathrm{x}=\mathrm{d} \mathrm{u}$, then
$\begin{aligned} \therefore \int \frac{1}{\sqrt{x+2}} d x &=\int \frac{1}{\sqrt{u}} d u=\int u^{-1 / 2} d u \\ &=\frac{u \frac{-1}{2}+1}{-1 / 2^{+1}}+C_{2} \quad \quad\left(\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right) \\ &=\frac{u \frac{-1}{2}+1}{-1 / 2^{+1}}+C_{2}=2 \sqrt{u}+C_{2} \\ &=2 \sqrt{x+2}+C_{2} \ldots \ldots . .(\mathrm{III}) \quad(\because u=x+2) \end{aligned}$
Also, $\int \frac{1}{(x-1) \sqrt{x+2}} d x$
Put $x+2=t^{2}=>d x=2 t d t$, then
$\begin{aligned} \int \frac{1}{(x-1) \sqrt{x+2}} \mathrm{dx} &=\int \frac{1}{\left(t^{2}-2-1\right) \sqrt{t^{2}}} 2 \mathrm{t} \mathrm{d} \mathrm{t} \\ &=2 \int \frac{1}{\left(t^{2}-3\right) t} \mathrm{t} \mathrm{d} \mathrm{t} \\ &=2 \int \frac{1}{t^{2}-(\sqrt{3})^{2}} \mathrm{~d} \mathrm{t} \end{aligned}$
$\begin{aligned} &=2 \cdot \frac{1}{2 . \sqrt{3}} \log \left|\frac{t-\sqrt{3}}{t+\sqrt{3}}\right|+C_{3} \quad\left(\because \int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C\right. \\ &=\frac{1}{\sqrt{3}} \log \left|\frac{\sqrt{x+2}-\sqrt{3}}{\sqrt{x+2}+\sqrt{3}}\right|+C_{3} \ldots .(\mathrm{IV}) \end{aligned}$
Put the values of equ.(II), (III) and (IV) in (I), then,
$\begin{aligned} \therefore & I=\frac{2}{3}(x+2)^{3 / 2}+C_{1}-\left(2 \sqrt{x+2}+C_{2}\right)+\frac{1}{\sqrt{3}} \log \left|\frac{\sqrt{x+2}-\sqrt{3}}{\sqrt{x+2}+\sqrt{3}}\right|+C_{1} \\ I &=\frac{2}{3}(x+2)^{3 / 2}-2 \sqrt{x+2}+\frac{1}{\sqrt{3}} \log \left|\frac{\sqrt{x+2}-\sqrt{3}}{\sqrt{x+2}+\sqrt{3}}\right|+C &\left(\because C=C_{1}-C_{2}+C_{3}\right) \text { Ans.. } \end{aligned}$

Indefinite Integrals Excercise 18.32 Question 5

Answer : - $2 \sqrt{x+1}+\frac{3}{2} \log\left| \frac{\sqrt{x+1}-2}{\sqrt{x+1}+2} \right|+C$
Hint :- Use substitution method and special integration formula to solve this problem..
Given :- $\int \frac{x}{(x-3) \sqrt{x+1}} d x$
Sol : - Let $I=\int \frac{x}{(x-3) \sqrt{x+1}} d x$
$\begin{aligned} &=\int \frac{x-3+3}{(x-3) \sqrt{x+1}} d x=\int \frac{(x-3)+3}{(x-3) \sqrt{x+1}} d x\\ &=\int\left\{\frac{x-3}{(x-3) \sqrt{x+1}}+\frac{3}{(x-3) \sqrt{x+1}}\right\} d x\\ &=\int\left\{\frac{1}{\sqrt{x+1}}+\frac{3}{(x-3) \sqrt{x+1}}\right\} d x\\ &=\int \frac{1}{\sqrt{x+1}} d x+\int \frac{3}{(x-3) \sqrt{x+1}} d x\\ &=\int \frac{1}{\sqrt{x+1}} d x+3 \int \frac{1}{(x-3) \sqrt{x+1}} d x \end{aligned}$
Now,$\int \frac{1}{\sqrt{x+1}} d x$
Let $x+1=p=>d x=d p$. Then,
$\begin{aligned} \therefore \int \frac{1}{\sqrt{x+1}} d x &=\int \frac{1}{\sqrt{p}} d p=\int p^{-1 / 2} \mathrm{dp} \\ &=\frac{p^{-1 / 2+1}}{-1 / 2^{+1}}+\mathrm{C}_{1} \quad\left(\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right) \\ &=\frac{p^{1 / 2}}{1 / 2}+\mathrm{C}_{1}=2 \sqrt{x+1}+\mathrm{C}_{1} \ldots \ldots_{1}(\mathrm{II}) \quad(\because \mathrm{p}=\mathrm{x}+1) \end{aligned}$

And, $\int \frac{1}{(x-3) \sqrt{x+1}} d x$

Put $\mathrm{x}+1=\mathrm{t}^{2}=>\mathrm{d} \mathrm{x}=2 \mathrm{t} \mathrm{d} \mathrm{t}$. Then,
$\begin{aligned} \therefore \int \frac{1}{(x-3) \sqrt{x+1}} d x=& \int \frac{1}{\left(t^{2}-1-3\right) \sqrt{t^{2}}} 2 \mathrm{t} \mathrm{d} \mathrm{t} \quad\left(\because \mathrm{x}=\mathrm{t}^{2}-1\right) \\ &=2 \int \frac{1}{\left(t^{2}-4\right) t} \mathrm{t} \mathrm{d} \mathrm{t} \\ &=2 \int \frac{1}{\left(t^{2}\right)-(2)^{2}} \mathrm{~d} \mathrm{t} \end{aligned}$
$\begin{array}{ll} =2 \cdot \frac{1}{2.2} \log \left|\frac{t-2}{t+2}\right|+\mathrm{C}_{2} & \left(\because \int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+\mathrm{C}\right) \\\\ \left.=\frac{1}{2} \log \left|\frac{\sqrt{x+1}-2}{\sqrt{x+1}+2}\right|+\mathrm{C}_{2} \ldots \ldots . \text { III }\right) & (\because \mathrm{t}=\sqrt{x+1}) \end{array}$
Put the values of equ. (II) and (III) in (I), Then ,
$\begin{aligned} &\mathrm{I}=2 \sqrt{x+1}+\mathrm{C}_{1}+3 \cdot \frac{1}{2} \log \left|\frac{\sqrt{x+1}-2}{\sqrt{x+1}+2}\right|+\mathrm{C}_{2} \\ &\mathrm{I}=2 \sqrt{x+1}+\frac{3}{2} \log \left|\frac{\sqrt{x+1}-2}{\sqrt{x+1}+2}\right|+\mathrm{C} \quad\left(\because \mathrm{C}=\mathrm{C}_{1}+\mathrm{C}_{2}\right) \text { Ans. } \end{aligned}$

Indefinite Integrals Excercise 18.32 Question 6

Answer : - $\frac{1}{\sqrt{2}} \tan ^{-1} \frac{x-1}{\sqrt{2 x}}--\frac{1}{2 \sqrt{2}} \log \left|\frac{x+1-\sqrt{2 x}}{x+1+\sqrt{2 x}}\right|+C$
Hint :- Use substitution method and special integration formula.
Given :-$\int \frac{1}{\left(x^{2}+1\right) \sqrt{x}} d x$
Sol : - Let $\mathrm{I}=\int \frac{1}{\left(x^{2}+1\right) \sqrt{x}} d x$
Put $\mathrm{x}=\mathrm{t}^{2}=>\mathrm{d} \mathrm{x}=2 \mathrm{t} \mathrm{d} \mathrm{t}$. then,
$\begin{aligned} \mathrm{I} &=\int \frac{1}{\left(\left(t^{2}\right)^{2}+1\right) \sqrt{t^{2}}} 2 t d t \\ &=\int \frac{1}{\left(t^{4}+1\right) t} 2 t d t \end{aligned}$
$=2 \int \frac{1}{\left(t^{4}+1\right)} d t=2 \int \frac{\frac{1}{t^{2}}}{\frac{t^{4}+1}{t^{2}}} d t$ (? dividing numerator and denominator by t2)

$\begin{aligned} &=2 \int \frac{\frac{1}{t^{2}}}{\frac{t^{4}}{t^{2}}+\frac{1}{t^{2}}} d t=2 \int \frac{\frac{1}{t^{2}}}{t^{2}+\frac{1}{t^{2}}} d t \\ &=\int \frac{\frac{1}{t^{2}}+\frac{1}{t^{2}}}{t^{2}+\frac{1}{t^{2}}} d t=\int \frac{\frac{1}{t^{2}}+1-1+\frac{1}{t^{2}}}{t^{2}+\frac{1}{t^{2}}} d t \\ &=\int \frac{\left(1+\frac{1}{t^{2}}\right)+\left(\frac{1}{t^{2}}-1\right)}{t^{2}+\frac{1}{t^{2}}} d t=\int \frac{\left(1+\frac{1}{t^{2}}\right)+\left(\frac{1}{t^{2}}-1\right)}{t^{2}+\frac{1}{t^{2}}} d t \end{aligned}$
$\begin{aligned} &=\int \frac{\left(1+\frac{1}{t^{2}}\right)}{t^{2}+\frac{1}{t^{2}}} d t-\int \frac{\left(1-\frac{1}{t^{2}}\right)}{t^{2}+\frac{1}{t^{2}}} d t \\ &=\int \frac{\left(1+\frac{1}{t^{2}}\right)}{t^{2}+\frac{1}{t^{2}}+2-2} d t-\int \frac{\left(1-\frac{1}{t^{2}}\right)}{t^{2}+\frac{1}{t^{2}}+2-2} d t \\ &\begin{aligned} =\int \frac{\left(1+\frac{1}{+t^{2}}\right)}{\left.\left(t-\frac{1}{t}\right)^{2}\right)+2} d t-\int \frac{\left(1-\frac{1}{t^{2}}\right)}{\left.\left(t+\frac{1}{t}\right)^{2}\right)-2} d t \quad \ldots \ldots \text { (I) } \quad\left(\because(a+b)^{2}=a^{2}+2 a b+b^{2}\right.\\ &\left.(a-b)^{2}=a^{2}-2 a b+b^{2}\right) \end{aligned} \end{aligned}$
$\begin{aligned} &\text { Now, } \int \frac{\left(1+\frac{1}{t^{2}}\right)}{\left.\left(t-\frac{1}{t}\right)^{2}+2\right)} d t \\ &\text { Put t }-1 / \mathrm{t}=\mathrm{u}=>1-\left(\frac{-1}{t^{2}}\right) \mathrm{dt}=\mathrm{d} \mathrm{u} \\ &\qquad \Rightarrow\left(1+\frac{1}{t^{2}}\right) \mathrm{d} \mathrm{t}=\mathrm{d} \mathrm{u} \text { then } \\ &\qquad \therefore \int \frac{\left(1+\frac{1}{t^{2}}\right)}{\left.\left(t-\frac{1}{t}\right)^{2}+2\right)} d t=\int \frac{d u}{\left.(u)^{2}+2\right)}=\int \frac{d u}{\left.(u)^{2}+(\sqrt{2})^{2}\right)} \end{aligned}$
$\begin{aligned} &=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{u}{\sqrt{2}}\right)+C_{1} \quad\left(\because \int \frac{d x}{a^{2}+x^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C\right) \\ &=\frac{1}{\sqrt{2}} \tan ^{-1} \frac{t-\frac{1}{t}}{\sqrt{2}}+C_{1} \quad(\because u=t-1 / t) \\ &=\frac{1}{\sqrt{2}} \tan ^{-1} \frac{t^{2}-1}{t}+C_{1} \\ &=\frac{1}{\sqrt{2}} \tan ^{-1} \frac{t^{2}-1}{\sqrt{2 t}}+C_{1} \ldots \ldots . . . \text { (II) } \end{aligned}$

And $\int \frac{\left(1-\frac{1}{t^{2}}\right)}{\left.\left(t+\frac{1}{t}\right)^{2}-2\right)} d t$

Put $\mathrm{t}+1 / \mathrm{t}=\mathrm{v} \quad=>\left(1-1 / \mathrm{t}^{2}\right) \mathrm{dt}=\mathrm{d} \mathrm{y}$ then,
$\begin{aligned} \therefore \int \frac{\left(1-\frac{1}{t^{2}}\right)}{\left.\left(t+\frac{1}{t}\right)^{2}-2\right)} d t &=\int \frac{d v}{\left.(v)^{2}+2\right)}=\int \frac{d v}{\left.(v)^{2}+(\sqrt{2})^{2}\right)} \\ &=\frac{1}{2 . \sqrt{2}} \log \left|\frac{v-\sqrt{2}}{v+\sqrt{2}}\right|+C \quad\left(\because \int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C\right) \\ &=\frac{1}{2 \cdot \sqrt{2}} \log \left|\frac{\left(t+\frac{1}{t}-\sqrt{2}\right)}{\left(t+\frac{1}{t}+\sqrt{2}\right)}\right|+C_{2} \end{aligned}$
$\begin{aligned} &=\frac{1}{2 . \sqrt{2}} \log \left|\frac{\left(t^{2}+1-\sqrt{2} t\right)}{\left(\frac{t^{2}+1+\sqrt{2} t}{t}\right)}\right|+C_{2} \\ &=\frac{1}{2 . \sqrt{2}} \log \left|\frac{\left(t^{2}+1-\sqrt{2} t\right)}{\left(t^{2}+1+\sqrt{2} t\right)}\right|+C_{2} \ldots \ldots . .(I I I) \end{aligned}$
Put the values of equ. (II) and (III) in equ.(I), We get,
$\begin{aligned} &I=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{\left(t^{2}-1\right)}{(\sqrt{2} t)}+C_{1}-\left(\frac{1}{2 \cdot \sqrt{2}} \log \left|\frac{\left(t^{2}+1-\sqrt{2} t\right)}{\left(t^{2}+1+\sqrt{2} t\right)}\right|+C_{2}\right)\right. \\ &I=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{\left(t^{2}-1\right)}{(\sqrt{2} t)}+C_{1}-\frac{1}{2 . \sqrt{2}} \log \left|\frac{\left(t^{2}+1-\sqrt{2} t\right)}{\left(t^{2}+1+\sqrt{2} t\right)}\right|-C_{2}\right) \\ &I=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{(x-1)}{(\sqrt{2} x)}-\frac{1}{2 . \sqrt{2}} \log \left|\frac{(x+1-\sqrt{2} x)}{(x+1+\sqrt{2} x)}\right|+C\right) \quad\left(\because t^{2}=x \text { and } c=c_{1}-c_{2}\right) \text { Ans.. } \end{aligned}$

Indefinite Integrals Excercise 18.32 Question 7

Answer : - $\frac{1}{\sqrt{2}} \log \left|\frac{x+2-\sqrt{2(x+1)}}{x+2+\sqrt{2(x+1)}}\right|+C$
Hint :- Use substitution and special integration formula.
Given :- $\int \frac{x}{\left(x^{2}+2 x+2\right) \sqrt{x+1}} d x$
Sol : -
$\begin{aligned} &\text { Let } \mathrm{I}=\int \frac{x}{\left(x^{2}+2 x+2\right) \sqrt{x+1}} d x \\ &\text { Put }, x+1=\mathrm{t}^{2} \\ &\mathrm{dx}=2 \mathrm{t} \mathrm{dt} \text { then } \end{aligned}$
$\begin{aligned} &\mathrm{I}=\int \frac{t^{2}-1}{\left\{\left(t^{2}-1\right)^{2}+2\left(t^{2}-1\right)+2\right\} \sqrt{t^{2}}} .2 \text { tdt }\left(\because t^{2}-1=x\right) \\ &=\int \frac{\left(t^{2}-1\right)}{\left(t^{4}+1-2 t^{2}+2 t^{2}-2+2\right) t} \cdot 2 \mathrm{t} \mathrm{dt} \\ &=2 \int \frac{t^{2}-1}{\left(t^{4}+1\right)} \mathrm{dt} \end{aligned}$
$=2 \int \frac{\frac{t^{2}-1}{t^{2}}}{\frac{t^{4}+1}{t^{2}}} \mathrm{dt}$ [Dividing num. And denom. By t2]
$\begin{aligned} &=2 \int \frac{t^{\frac{2}{2}}-\frac{1}{t^{2}}}{\frac{t^{4}}{t^{2}}+\frac{1}{t^{2}}} \mathrm{dt} \\ &=2 \int \frac{\left(1-\frac{1}{t^{2}}\right)}{t^{2}+\frac{1}{t^{2}}} \mathrm{dt} \end{aligned}$
$\begin{aligned} &=2 \int \frac{\left(1-\frac{1}{t^{2}}\right)}{t^{2}+\frac{1}{t^{2}}+2-2} d t \\ &=2 \int \frac{\left(1-\frac{1}{t^{2}}\right)}{\left(t^{2}+\frac{1}{t^{2}}+2 . t_{t}^{1}\right)-2} d t \\ &=2 \int \frac{\left(1-\frac{1}{t^{2}}\right)}{\left(t+\frac{1}{t}\right)^{2}-2} \text { dt }\left[\because(a+b)^{2}=a^{2}+2 a b+b^{2}\right] \end{aligned}$
$\begin{aligned} &\text { Again, put } \mathrm{t}+\frac{1}{t}=\mathrm{u} \\ &\Rightarrow\left(1-\frac{1}{t^{2}}\right) d t=\mathrm{du} \text { then } \\ &\qquad \mathrm{I}=2 \int \frac{d u}{u^{2}-(\sqrt{2})^{2}} \end{aligned}$

$\begin{aligned} &=2 \cdot \frac{1}{2 \cdot \sqrt{2}} \log \left|\frac{u-\sqrt{2}}{u+\sqrt{2}}\right|+c \quad\left[\because \int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+c\right]\\ &=\frac{1}{\cdot \sqrt{2}} \log \left|\frac{t+\frac{1}{t}-\sqrt{2}}{t+\frac{1}{t}+\sqrt{2}}\right|+c \quad\left[\because \mathrm{u}=\mathrm{t}+\frac{1}{t}\right]\\ &=\frac{1}{\sqrt{2}} \log \left|\frac{\frac{t^{2}+1-\sqrt{2} t}{t}}{\frac{t^{2}+1+\sqrt{2} t}{t}}\right|+c \end{aligned}$

$\begin{aligned} &=\frac{1}{\sqrt{2}} \log \left|\frac{t^{2}+1-\sqrt{2} t}{t^{2}+1+\sqrt{2} t}\right|+\mathrm{c} \\ &=\frac{1}{\cdot \sqrt{2}} \log \left|\frac{x+1+1-\sqrt{2} \cdot \sqrt{x+1}}{x+1+1+\sqrt{2} \cdot \sqrt{x+1}}\right|+\mathrm{c} \quad\left[\because \mathrm{t}=\sqrt{x+1} \text { or } t^{2}=x+1\right] \\ &=\frac{1}{\sqrt{2}} \log \left|\frac{x+2-\cdot \sqrt{2(x+1)}}{x+2+\sqrt{2(x+1)}}\right|+\mathrm{c} \end{aligned}$

Indefinite Integrals Excercise 18.32 Question 8

Answer : -$-\frac{1}{\sqrt{2}} \log \left|\left(\frac{1}{x-1}+\frac{1}{2}\right)+\sqrt{\left(\frac{1}{x-1}+\frac{1}{2}\right)^{2}+\frac{1}{4}}\right|+C$
Hint :- Use substitution and special integration formula.
Given :- $\int \frac{1}{\left ( x-1 \right )\sqrt{x^{2}+1}}dx$
Sol : - Let $I = \int \frac{1}{\left ( x-1 \right )\sqrt{x^{2}+1}}dx$
Put, $x -1 = \frac{1}{t}$
$\Rightarrow dx=-\frac{1}{t^{2}}dt$ then,
$\begin{aligned} &\begin{aligned} \mathrm{I} &=\int \frac{1}{\frac{1}{t} \sqrt{\left(\frac{1}{t}+1\right)^{2}+1}}\left(-\frac{1}{t^{2}}\right) d t \\ &=-\int \frac{\frac{1}{t^{2}}}{\frac{1}{t} \sqrt{\frac{1}{t^{2}}+1+\frac{3}{t}+1}} d t\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right] \end{aligned} \end{aligned}$
$\begin{aligned} &=-\int \frac{\frac{1}{t^{2}}}{\frac{1}{t}\sqrt{\frac{1+t^{2}+2t+t^{2}}{t^{2}}}}dt \\ &=-\int \frac{\frac{1}{t^{2}}}{\frac{11}{t} \cdot \frac{1}{t} \sqrt{2 t^{2}+2 t+1}} d t \\ &=-\int \frac{\frac{1}{t^{2}}}{\frac{1}{t^{2}} \sqrt{2\left(t^{2}+t+\frac{1}{2}\right)}} d t \end{aligned}$
$\begin{aligned} &=-\int \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{t^{2}+2 \cdot t \cdot \frac{1}{2}+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}+\frac{1}{2}}} d t \\ &=-\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{\left(t+\frac{1}{2}\right)^{2}-\frac{1}{4}+\frac{1}{2}}} d t&\quad\left[\because a^{2}+b^{2}+2 a b=(a+b)^{2}\right] \\ \end{aligned}$
$\begin{aligned} &=-\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{\left(t+\frac{1}{2}\right)^{2}-\left(\frac{1-2}{4}\right)}} d t \\ &=-\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{\left(t+\frac{1}{2}\right)^{2}-\left(\frac{-1}{4}\right)}} d t \\ &=-\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{\left(t+\frac{1}{2}\right)^{2}+\frac{1}{4}}} d t \\ &=-\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{\left(t+\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}}} d t \end{aligned}$
Put, $t +\frac{1}{2} = u$

$\Rightarrow dt=du$ than

$\begin{aligned} &\mathrm{I}=-\frac{1}{2} \int \frac{1}{\sqrt{u^{2}+\left(\frac{1}{2}\right)^{2}}} \mathrm{du} \\ &\mathrm{I}=-\frac{1}{\sqrt{2}} \log \left|u+\sqrt{u^{2}+\left(\frac{1}{2}\right)^{2}}\right|+\mathrm{c} \quad\left[\because \int \frac{1}{\sqrt{x^{2}+a^{2}}} d x=\log \left|x+\sqrt{x^{2}+a^{2}}\right|+\mathrm{c}\right] \end{aligned}$
$\begin{aligned} &\mathrm{I}=-\frac{1}{\sqrt{2}} \log \left|\left(t+\frac{1}{2}\right)+\sqrt{\left(t+\frac{1}{2}\right)^{2}+\frac{1}{4}}\right|+\mathrm{c} \quad\left(\because \mathrm{u}=\mathrm{t}+\frac{1}{2}\right) \\ &\mathrm{I}=-\frac{1}{\sqrt{2}} \log \left|\left(\frac{1}{x-1}+\frac{1}{2}\right)+\sqrt{\left(\frac{1}{x-1}+\frac{1}{2}\right)^{2}+\frac{1}{4}}\right|+\mathrm{c} \quad\left(\because(\mathrm{x}-1)=\frac{1}{t}\right) \end{aligned}$

Indefinite Integrals Excercise 18.32 Question 9

Answer : - $-\log \left|\left(\frac{1}{x+1}-\frac{1}{2}\right)+\sqrt{\left(\frac{1}{x+1}-\frac{1}{2}\right)^{2}+\frac{3}{4}}\right|+C$
Hint :- Use substitution method and special integration formula to solve this integration
Given :- $\int \frac{1}{(x+1) \sqrt{x^{2}+x+1}} d x$
Sol : - Let $\mathrm{I}=\int \frac{1}{(x+1) \sqrt{x^{2}+x+1}} d x$
Put, $x + 1 = \frac{1}{t}$
$\Rightarrow dx=-\frac{1}{t^{2}}dt$ then,
$\begin{aligned} &\mathrm{I}=\int \frac{1}{\frac{1}{t} \sqrt{\left(\frac{1}{t}-1\right)^{2}+\left(\frac{1}{t}-1\right)+1}}\left(-\frac{1}{t^{2}}\right) d t\left(\because \mathrm{x}=\frac{1}{t}-1\right) \\ &=-\int \frac{\frac{1}{t^{2}}}{\frac{1}{t} \sqrt{\frac{1}{t^{2}}+1-\frac{2}{t}+\frac{1}{t}-1+1}} d t \quad \because(a-b)^{2}=a^{2}-2 a b+b^{2} \\ &=-\int \frac{\frac{1}{t^{2}}}{\frac{1}{t} \sqrt{t^{2}-\frac{1}{t}+1}} d t \end{aligned}$
$\begin{aligned} &=-\int \frac{\frac{1}{t^{2}}}{\frac{1}{t} \sqrt{\frac{1-t+t^{2}}{t^{2}}}} d t \\ &=-\int \frac{\frac{1}{t^{2}}}{\frac{1}{t}^{2} \bar{t} \sqrt{1-t+t^{2}}} d t \\ &=-\int \frac{\frac{1}{t^{2}}}{\frac{1}{t^{2}} \sqrt{t^{2}-t+1}} d t \\ &=-\int \frac{1}{\sqrt{t^{2}-2 \cdot t_{2}^{\frac{1}{2}}+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}+1}} d t \end{aligned}$

$\begin{aligned} &=-\int \frac{1}{\sqrt{\left(t-\frac{1}{2}\right)^{2}-\frac{1}{4}+1}} d t \\ &=-\int \frac{1}{\sqrt{\left(t-\frac{1}{2}\right)^{2}-\left(\frac{1-4}{4}\right)}} d t \end{aligned}$

$\begin{aligned} &=-\int \frac{1}{\sqrt{\left(t-\frac{1}{2}\right)^{2}-\left(\frac{-3}{4}\right)}} d t \\ &=-\int \frac{1}{\sqrt{\left(t-\frac{1}{2}\right)^{2}+\left(\frac{3}{4}\right)}} d t \\ &=-\int \frac{1}{\sqrt{\left(t-\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}} d t \end{aligned}$
Put,$t-\frac{1}{2}=u$
$\Rightarrow dt=du$ than,
$\begin{aligned} &\mathrm{I}=-\int \frac{1}{\sqrt{u^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}} d u \\ &=-\log \left|u+\sqrt{u^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}\right|+\mathrm{c} \quad\left[\because \int \frac{1}{\sqrt{x^{2}+a^{2}}} d x=\log \left|x+\sqrt{x^{2}+a^{2}}\right|+\mathrm{c}\right] \end{aligned}$
$\begin{aligned} &=-\log \left|\left(t-\frac{1}{2}\right)+\sqrt{\left(t-\frac{1}{2}\right)^{2}+\frac{3}{4}}\right|+\mathrm{c} \quad\left(\because \mathrm{u}=\mathrm{t}-\frac{1}{2}\right) \\ &=-\log \left|\left(\frac{1}{x+1}-\frac{1}{2}\right)+\sqrt{\left(\frac{1}{x+1}-\frac{1}{2}\right)^{2}+\frac{3}{4}}\right| \mid+c \quad\left[\because \mathrm{t}=\frac{1}{x+1}\right] \end{aligned}$

Indefinite Integrals Excercise 18.32 Question 10

Answer : - $-\frac{1}{2 \sqrt{2}} \log \left|\frac{\sqrt{2} x+\sqrt{x^{2}+1}}{\sqrt{2} x-\sqrt{x^{2}+1}}\right|+C$
Hint :- Use substitution method and special integration formula.
Given :- $\int \frac{1}{\left(x^{2}-1\right) \sqrt{x^{2}+1}} d x$
Sol : - Let $\mathrm{I}=\int \frac{1}{\left(x^{2}-1\right) \sqrt{x^{2}+1}} d x$
Put, $x=\frac{1}{t}$
$\Rightarrow dx=-\frac{1}{t^{2}}dt$ then,
$\begin{aligned} &\quad I=\int \frac{1}{\left(\frac{1}{t^{2}}-1\right) \sqrt{\frac{1}{t^{2}}+1}} \cdot\left(-\frac{1}{t^{2}}\right) \operatorname{dt}\left(\because x=\frac{1}{t}\right) \\ &I=-\int \frac{1}{\left(\frac{1-t^{2}}{t^{2}}\right) \sqrt{\frac{1+t^{2}}{t^{2}}}} \cdot \frac{1}{t^{2}} d t \end{aligned}$
$\begin{aligned} &I=-\int \frac{\frac{1}{t^{2}}}{\frac{1}{t^{2}}\left(1-t^{2}\right) \cdot \frac{1}{t}, \sqrt{1+t^{2}}} d t \\ &I=-\int \frac{\frac{1}{t^{2}}}{\left(1-t^{2}\right), \frac{1}{t} \cdot \sqrt{1+t^{2}}} d t \\ &I=-\int \frac{t d t}{\left(1-t^{2}\right) \sqrt{1+t^{2}}} \end{aligned}$

$\begin{aligned} &\text { Put, } 1+t^{2}=u^{2} \\ &\Rightarrow 2 \mathrm{t} \mathrm{dt}=2 \mathrm{u} \mathrm{du} \\ &\mathrm{I}=-\int \frac{1}{\left\{1-\left(u^{2}-1\right)\right] \sqrt{u^{2}}} \cdot \mathrm{u} \mathrm{d} \mathrm{u} \\ &\mathrm{I}=-\int \frac{1}{\left(1-u^{2}+1\right) u} \cdot \mathrm{u} \mathrm{du} \\ &\mathrm{I}=-\int \frac{1}{2-u^{2}} \mathrm{du} \\ &\mathrm{I}=-\int \frac{1}{(\sqrt{2})^{2}-u^{2}} \mathrm{du} \end{aligned}$

$\begin{array}{ll} \mathrm{I}=-\frac{1}{2 \sqrt{2}} \log \left|\frac{\sqrt{2}+u}{\sqrt{2}-u}\right|+\mathrm{c} \quad & {\left[\because \int \frac{1}{a^{2}-x^{2}} d x=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+\mathrm{c}\right]} \\\\ \mathrm{I}=-\frac{1}{2 \sqrt{2}} \log \left|\frac{\sqrt{2}+\sqrt{1+t^{2}}}{\sqrt{2}-\sqrt{1+t^{2}}}\right|+\mathrm{c} \quad & {\left[\because 1+t^{2}=u^{2} \text { or } u=\sqrt{1+t^{2}}\right]} \end{array}$
$\begin{aligned} &\mathrm{I}=-\frac{1}{2 \sqrt{2}} \log \left|\frac{\sqrt{2}+\sqrt{1+\frac{1}{x^{2}}}}{\sqrt{2}-\sqrt{1+\frac{1}{x^{2}}}}\right|+\mathrm{c} \\ &\mathrm{I}=-\frac{1}{2 \sqrt{2}} \log \left|\frac{\sqrt{2}+\sqrt{\frac{x^{2}+1}{x^{2}}}}{\sqrt{2}-\sqrt{\frac{x^{2}+1}{x^{2}}}}\right|+\mathrm{c} \quad \quad\left[\because \mathrm{t}=\frac{1}{x}\right] \end{aligned}$
$\begin{aligned} &I=-\frac{1}{2 \sqrt{2}} \log \left|\frac{\sqrt{2}+\frac{\sqrt{x^{2}+1}}{x}}{\sqrt{2}-\frac{\sqrt{x^{2}+1}}{x}}\right|+c \\ &I=-\frac{1}{2 \sqrt{2}} \log \left|\frac{\frac{\sqrt{2} x+\sqrt{x^{2}+1}}{x}}{\frac{\sqrt{2} x-\sqrt{x^{2}+1}}{x}}\right|+c \\ &I=-\frac{1}{2 \sqrt{2}} \log \left|\frac{\sqrt{2} x+\sqrt{x^{2}+1}}{\sqrt{2} x-\sqrt{x^{2}+1}}\right|+c \end{aligned}$

Indefinite Integrals Excercise 18.32 Question 11

Answer : - $\frac{1}{\sqrt{3}} \tan ^{-1}\left(\sqrt{\frac{x^{2}+1}{3}}\right)+C$
Hint :- Use substitution method and special integration formula..
Given :- $\int \frac{x}{\left(x^{2}+4\right) \sqrt{x^{2}+1}} d x$
Sol : - Let $\mathrm{I}=\int \frac{x}{\left(x^{2}+4\right) \sqrt{x^{2}+1}} d x$
$\begin{array}{r} \text { Put } x^{2}+1=t^{2} \Rightarrow 2 \mathrm{xd} \mathrm{x}=2 \mathrm{t} \mathrm{dt} \rightarrow \mathrm{x} \mathrm{dx}=\mathrm{t} \mathrm{dt} \text { than, } \\\\ \mathrm{I}=\int \frac{1}{\left(t^{2}+3\right) \sqrt{t^{2}}} \mathrm{t} \mathrm{dt}\left(\because x^{2}+1=\mathrm{t}^{2} \rightarrow x^{2}=t^{2}-1\right. \\\\ \left.\rightarrow x^{2}+4=t^{2}-1+4 \rightarrow x^{2}+4=t^{2}+3\right) \end{array}$
$\begin{aligned} &\mathrm{I}=\int \frac{1}{\left(t^{2}+3\right) \cdot t} \mathrm{t} \mathrm{dt} \\ &\mathrm{I}=\int \frac{1}{t^{2}+3} \mathrm{dt} \\ &\mathrm{I}=\int \frac{1}{t^{2}+(\sqrt{3})^{2}} \mathrm{dt} \end{aligned}$
$\begin{aligned} &I=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{t}{\sqrt{2}}\right)+c \quad\left[\because \int \frac{1}{a^{2}+x^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c\right] \\ &I=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{\sqrt{x^{2}+1}}{\sqrt{3}}\right)+c \\ &I=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\sqrt{\frac{x^{2}+1}{3}}\right)+c \quad\left[\because t=\sqrt{\left.x^{2}+1\right]}\right. \end{aligned}$

Indefinite Integrals Excercise 18.32 Question 12

Answer : - $-\frac{1}{\sqrt{2}} \tan ^{-1}\left(\sqrt{\frac{1-x^{2}}{2 x^{2}}}\right)+C$
Hint :- Use substitution method and special integration formula..
Given :-$\int \frac{1}{\left(x^{2}+1\right) \sqrt{1-x^{2}}} d x$
Sol : - Let $\mathrm{I}=\int \frac{1}{\left(x^{2}+1\right) \sqrt{1-x^{2}}} d x$
Put $\mathrm{x}=\frac{1}{t} \rightarrow \mathrm{d} \mathrm{x}=-\frac{1}{t^{2}} \mathrm{dt}$ then,
$\begin{aligned} I &=\int \frac{1}{\left(1+\frac{1}{t^{2}}\right) \sqrt{1-\frac{1}{t^{2}}}}\left(-\frac{1}{t^{2}}\right) d t \\ &=-\int \frac{\frac{1}{t^{2}}}{\left(\frac{t^{2}+1}{t^{2}}\right) \sqrt{\frac{t^{2}-1}{t^{2}}}} d t \\ &=-\int \frac{\frac{1}{t^{2}}}{\frac{1}{t^{2}}\left(t^{2}+1\right) \cdot \frac{1}{t} \sqrt{t^{2}-1}} d t \end{aligned}$
$=-\int \frac{t}{\left(t^{2}+1\right) \sqrt{t^{2}-1}} \mathrm{dt}$
Put, $t^{2}-1=u^{2} \rightarrow 2 \mathrm{t} \mathrm{dt}=2 \mathrm{u} \mathrm{du} \rightarrow \mathrm{t} \mathrm{dt}=\mathrm{u} \mathrm{du}$ than,
$\begin{aligned} &\quad I=-\int \frac{u d u}{\left(u^{2}+1+1\right) \sqrt{u^{2}}} \quad\left[\because t^{2}-1=u^{2} \rightarrow t^{2}=u^{2}+1\right] \\ &=-\int \frac{u d u}{\left(u^{2}+2\right) u} \\ &\quad=-\int \frac{d u}{u^{2}+(\sqrt{2})^{2}} \end{aligned}$
$\begin{aligned} &=-\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{u}{\sqrt{2}}\right)+\mathrm{c} \quad\left[\because \int \frac{1}{a^{2}+x^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+\mathrm{c}\right] \\ &=-\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{\sqrt{t^{2}-1}}{\sqrt{2}}\right)+\mathrm{c} \quad\left[\because u^{2}=t^{2}-1 \rightarrow \mathrm{u}=\sqrt{t^{2}-1}\right] \\ &=-\frac{1}{\sqrt{2}} \tan ^{-1}\left(\sqrt{\frac{t^{2}-1}{2}}\right)+\mathrm{c} \end{aligned}$
$\begin{aligned} &=-\frac{1}{\sqrt{2}} \tan ^{-1}\left(\sqrt{\frac{\frac{1}{x^{2}}-1}{2}}\right)+\mathrm{c} \quad\left[\because \mathrm{x}=\frac{1}{t} \rightarrow \mathrm{t}=\frac{1}{x}\right]\\ &=-\frac{1}{\sqrt{2}} \tan ^{-1}\left(\sqrt{\frac{\frac{1-x^{2}}{x^{2}}}{2}}\right)+\mathrm{c}\\ &=-\frac{1}{\sqrt{2}} \tan ^{-1}\left(\sqrt{\frac{1-x^{2}}{2 x^{2}}}\right)+\mathrm{c} \end{aligned}$

Indefinite Integrals Excercise 18.32 Question 13

Answer : - $\frac{1}{2 \sqrt{33}} \log \left|\frac{\sqrt{11} x+\sqrt{3 x^{2}-12}}{\sqrt{11} x-\sqrt{3 x^{2}-12}}\right|+C$
Hint :- Use substitution method and special integration formula.
Given :- $\int \frac{1}{\left(2 x^{2}+3\right) \sqrt{x^{2}-4}} d x$
Sol : -
$\begin{aligned} &\text { Let } \mathrm{I}=\int \frac{1}{\left(2 x^{2}+3\right) \sqrt{x^{2}-4}} d x \\ &\text { let, } \mathrm{x}=\frac{1}{t} \\ & \mathrm{dx}=-\frac{1}{t^{2}} \mathrm{dt} \text { then } \end{aligned}$
$\begin{gathered} \quad I=\int \frac{1}{\left(2 \cdot \frac{1}{t^{2}}+3\right) \sqrt{\frac{1}{t^{2}}-4}}\left(-\frac{1}{t^{2}} \mathrm{dt}\right) \\ =-\int \frac{1}{\left(\frac{2+3 t^{2}}{t^{2}}\right) \sqrt{\frac{1-4 t^{2}}{t^{2}}}} \cdot \frac{1}{t^{2}} \mathrm{dt} \\ =-\int \frac{\frac{1}{t^{2}}}{\frac{1}{t^{2}}\left(3 t^{2}+2\right) \sqrt{1-4 t^{2}} \cdot \frac{1}{t}} \mathrm{dt} \\ \quad=-\int \frac{t}{\left(3 t^{2}+2\right) \sqrt{1-4 t^{2}}} \mathrm{dt} \end{gathered}$
$\begin{aligned} &\text { Put } \text { . } 1-4 t^{2}=p^{2} \\ &\Rightarrow-4.2 \mathrm{t} \mathrm{dt}=2 \mathrm{p} \mathrm{dp} \\ &\Rightarrow \mathrm{tdt}=-\frac{p d p}{4} \end{aligned}$
$\mathrm{I}=-\int \frac{1}{\left\{3\left(\frac{1-p^{2}}{4}\right)+2\right\} \cdot \sqrt{p^{2}}} \cdot-\frac{p d p}{4}\left[\because 1-4 t^{2}=p^{2} \rightarrow 1-p^{2}=4 t^{2} \rightarrow t^{2}=\frac{1-p^{2}}{4}\right]$
$\begin{aligned} &\mathrm{I}=\frac{1}{4} \int \frac{1}{\frac{3-3 p^{2}+8}{4} \cdot p} \cdot \mathrm{pdp} \\ &\mathrm{I}=\frac{1}{4} \int \frac{4}{11-3 p^{2}} \mathrm{dp} \\ &\mathrm{I}=\frac{1}{3} \int \frac{1}{\left(\frac{11}{3}-p^{2}\right)} \mathrm{dp} \end{aligned}$
$\begin{aligned} &\mathrm{I}=\frac{1}{3} \int \frac{1}{\left(\sqrt{\frac{11}{3}}\right)^{2}-p^{2}} \mathrm{dp} \\ &\mathrm{I}=\frac{1}{3} \frac{1}{2 \cdot \frac{\sqrt{11}}{\sqrt{3}}} \log \left|\frac{\sqrt{\frac{11}{3}}+p}{\sqrt{\frac{11}{3}}-p}\right|+\mathrm{c}\left[\because \int \frac{1}{a^{2}-x^{2}} d x=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|\right. \end{aligned}$

$\begin{aligned} &=\frac{1}{2 \sqrt{3} \cdot \sqrt{11}} \log \left|\frac{\sqrt{\frac{11}{3}}+\sqrt{1-4 t^{2}}}{\sqrt{\frac{11}{3}}-\sqrt{1-4 t^{2}}}\right|+c\left[\because p^{2}=1-4 t^{2} \rightarrow \mathrm{p}=\sqrt{1-4 t^{2}}\right]\\ &=\frac{1}{2 \sqrt{33}} \log \left|\frac{\frac{\sqrt{11}+\sqrt{3} \sqrt{1-4 t^{2}}}{\sqrt{3}}}{\frac{\sqrt{11}-\sqrt{3} \sqrt{1-4 t^{2}}}{\sqrt{3}}}\right|+c \end{aligned}$

$\begin{aligned} &=\frac{1}{2 \sqrt{33}} \log \left|\frac{\sqrt{11}+\sqrt{3\left(1-4 t^{2}\right)}}{\sqrt{11}-\sqrt{3\left(1-4 t^{2}\right)}}\right|+\mathrm{c} \\ &=\frac{1}{2 \sqrt{33}} \log \left|\frac{\sqrt{11}+\sqrt{3\left(1-4 \frac{1}{x^{2}}\right)}}{\sqrt{11}-\sqrt{3\left(1-4 \frac{1}{x^{2}}\right)}}\right|+\mathrm{c}\left[\because \mathrm{x}=\frac{1}{t} \rightarrow \mathrm{t}=\frac{1}{\mathrm{~m}}\right] \end{aligned}$
$\begin{aligned} &=\frac{1}{2 \sqrt{33}} \log \left|\frac{\sqrt{11}+\sqrt{\frac{3 x^{2}-12}{x^{2}}}}{\sqrt{11}-\sqrt{\frac{3 x^{2}-12}{x^{2}}}}\right|+\mathrm{c} \\ &=\frac{1}{2 \sqrt{33}} \log \left|\frac{\sqrt{11}+\frac{1}{x} \sqrt{3 x^{2}-12}}{\sqrt{11}-\frac{1}{x} \sqrt{3 x^{2}-12}}\right|+\mathrm{c} \end{aligned}$
$\begin{aligned} &=\frac{1}{2 \sqrt{33}} \log \left|\frac{\frac{\sqrt{11} x+\sqrt{3 x^{2}-12}}{x}}{\frac{\sqrt{\frac{11}{x}-\sqrt{3 x^{2}-12}}}{x}}\right|+\mathrm{c}\\ &=\frac{1}{2 \sqrt{33}} \log \left|\frac{\sqrt{11} x+\sqrt{3 x^{2}-12}}{\sqrt{11} x-\sqrt{3 x^{2}-12}}\right|+\mathrm{c} \end{aligned}$

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Indefinite Integrals Excercise 18.32 Question 14

Answer : - $\frac{1}{2 \sqrt{5}} \log \left|\frac{\sqrt{x^{2}+9}-\sqrt{5}}{\sqrt{x^{2}+9}+\sqrt{5}}\right|+C$
Hint :- Use substitution method and special integration formula.
Given :- $\int \frac{x}{\left(x^{2}+4\right) \sqrt{x^{2}+9}} d x$
Sol : -
$\begin{aligned} &\text { Let } \mathrm{I}=\int \frac{x}{\left(x^{2}+4\right) \sqrt{x^{2}+9}} d x \\ &\text { put, } x^{2}+9=u^{2} \rightarrow 2 \mathrm{x} \mathrm{dx}=2 \mathrm{u} \mathrm{du} \\ &\quad \rightarrow \mathrm{x} \mathrm{dx}=\mathrm{u} \text { du than } \\ &\qquad \mathrm{I}=\int \frac{u d u}{\left(u^{2}-9+4\right) \sqrt{u^{2}}} \end{aligned}$
$\begin{aligned} &\mathrm{I}=\int \frac{u d u}{\left(u^{2}-5\right) u} \quad\left[\because x^{2}+9=u^{2} \rightarrow x^{2}=u^{2}-9\right] \\ &\mathrm{I}=\int \frac{u}{u^{2}-5} \mathrm{~d} \mathrm{u} \\ &\mathrm{I}=\int \frac{u}{u^{2}-(\sqrt{5})^{2}} \mathrm{~d} \mathrm{u} \end{aligned}$
$\begin{aligned} &\mathrm{I}=\frac{1}{2 \cdot \sqrt{5}} \log \left|\frac{u-\sqrt{5}}{u+\sqrt{5}}\right|+\mathrm{c} \quad\left[\because \int \frac{1}{x^{2}-a^{2}} \mathrm{dx}=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+c\right] \\ &\mathrm{I}=\frac{1}{2 \sqrt{5}} \log \left(\frac{\sqrt{x^{2}+9}-\sqrt{5}}{\sqrt{x^{2}+9}+\sqrt{5}}\right) +\mathrm{C} \quad\left[\ddot x^{2}+9=u^{2} \rightarrow \mathrm{u}=\sqrt{x^{2}+9}\right] \end{aligned}$

Class 12 RD Sharma chapter 18 exercise 18.32 solution is prepared specifically for students who want to get good insight on the subject. It contains questions and answers in the same place making it easier for students to prepare and revise. Class 12 RD Sharma chapter 18 exercise 18.32 solution is a material that follows the CBSE syllabus and is updated to the latest edition.

RD Sharma Class 12th exercise 18.32 answers that are immediate and assessed. It further assists students with understanding formulae and keeping an eye on methods. RD Sharma class 12 chapter 18 exercise 18.32 has around 14 questions, which have subparts. Given below are some of the topics covered:

Some huge integrals close by theorems
A blend of sensible mathematical limits by using inadequate parts
Exactly when the denominator contains some reiterating straight factors, then the denominator contains unchangeable quadratic factors.
Blend of some interesting absurd arithmetical limits

Given below are the benefits of using Rd Sharma Class 12th exercise 18.32 material:

Experts have given these solutions in a basic and straightforward way to empower students to adapt appropriately.
These solutions are helpful for finishing schoolwork and reading for CBSE examinations as they agree with the CBSE prospectus.
These solutions additionally help free comprehension of thoughts and getting ready for 12th sheets as well as forthcoming serious exams like BITSAT, JEE, and NEET.
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