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

Edited By Kuldeep Maurya | Updated on Jan 24, 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.

## 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}

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

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RD Sharma Chapter-wise Solutions

1. How many questions are there in Class 12 RD Sharma chapter 18?

There are a total of 14 questions in RD Sharma Solutions Class 12 RD Sharma chapter 18 exercise 18.32.

2. . How many exercises are there in Chapter 18?

There are a total of 32 exercises, i.e., from 18.1 to 18.32.

3. How can I access the solution of RD Sharma chapter 18 exercise 18.30 for class 12?

Class 12 RD Sharma Chapter 18 Exercise 18.30 Solution is available on Career360's website and is free to access.

4. Do RD Sharma Relations Ex 18.30 class 12 solutions follow the latest version of the book?

Yes, the Class 12 RD Sharma Chapter 18 Exercise 18.4 Solution is updated to the latest version.

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