# Get Answers to all your Questions

#### Need solution for RD Sharma Maths Class 12 Chapter 18 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}