Provide Solution For  R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise  Revision Exercise Question 97 Maths Textbook Solution.

$x \cdot \log \left(x+\sqrt{x^{2}+a^{2}}\right)-\sqrt{x^{2}-a^{2}}+C\left[\because t=x^{2}+a^{2}\right]$

Hint

$\text { Put } x=a \tan \theta$

Given:

$\int \log \left(x+\sqrt{x^{2}+a^{2}}\right) d x$

Solution:

$I=\int 1_{I I} \cdot \log \left(x+\sqrt{x_{I}^{2}+a^{2}}\right) d x$

$=\log \left(x+\sqrt{x^{2}+a^{2}}\right) \int 1 d x-\int\left[\frac{d}{d x}\left\{\log \left(x+\sqrt{x^{2}+a^{2}}\right)\right\} \int 1 d x\right]$

$=\log \left(x+\sqrt{x^{2}+a^{2}}\right) \cdot x-\int\left(\frac{1}{x+\sqrt{x^{2}+a^{2}}}\right) \times\left(1+\frac{1 \times 2 x}{2 \sqrt{x^{2}+a^{2}}}\right) \cdot x \cdot d x$

$=\log \left(x+\sqrt{x^{2}+a^{2}}\right) \cdot x-\int \frac{x}{\sqrt{x^{2}+a^{2}}} d x$

$\text { putting } x^{2}+a^{2}=\text { tinthe sec o nd inte gral }$

$\Rightarrow 2 x d x=d t$

$\Rightarrow x d x=\frac{d t}{2}$

$I=x \cdot \log \left(x+\sqrt{x^{2}+a^{2}}\right)-\frac{1}{2} \int \frac{1}{\sqrt{t}} d t$

$=x \cdot \log \left(x+\sqrt{x^{2}+a^{2}}\right)-\frac{1}{2} \int t^{-\frac{1}{2}} d t$

$=x \cdot \log \left(x+\sqrt{x^{2}+a^{2}}\right)-\frac{1}{2}\left[\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\right]+C$

$=x \cdot \log \left(x+\sqrt{x^{2}+a^{2}}\right)-\sqrt{t}+C$

$=x \cdot \log \left(x+\sqrt{x^{2}+a^{2}}\right)-\sqrt{x^{2}-a^{2}}+C\left[\because t=x^{2}+a^{2}\right]$