Question

Prove that :-
$\sqrt{x^2+a^2}dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log \left|x+\sqrt{x^2+a^2}\right|+c$

Answer 1

$\int \sqrt{x^2+a^2}dx$

$x=a\tan \theta ,\tan \theta =\frac{x}{a}$

$dx=a{\sec }^2\theta \alpha \theta$

$\int \sqrt{a^2\left(1+{\tan }^2\theta \right)}.a{\sec }^2\theta \alpha \theta$

$\int a\sec \theta .a{\sec }^2\theta \alpha \theta$

$=a^2\int \sec \theta \alpha \theta$

$=a^2\left[\frac{\sec \theta \tan \theta }{2}+\frac{1}{2}\int \sec \theta \alpha \theta \right]$ (Reduction formula)

$=\frac{a^2}{2}\sec \theta \tan \theta +\frac{a^2}{2}\log \left|\sec \theta +\tan \theta \right|$

$=\frac{a^2}{2}\frac{\sqrt{x^2+a^2}}{a}.\frac{x}{a}+\frac{a^2}{2}\log \left|\frac{\sqrt{x^2+a^2}}{a}+\frac{x}{a}\right|$

$=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log \left|x+\sqrt{x^2+a^2}\right|+c$