Question

$\int \frac{1}{\sqrt{x^2+a^2}}dx=log|x+\sqrt{x^2+a^2}|+c$

Answer 1

LHS.

We have,

$\int \frac{dx}{\sqrt{x^2-a^2}}$

We will use put

$x=a\sec \theta$

Thus, on differentiating and we get,

$dx=a\sec \theta \tan \theta d\theta$

Then,

$\int \frac{a\sec \theta \tan \theta d\theta }{\sqrt{a^2{\sec }^2\theta -a^2}}$

$\Rightarrow \int \frac{a\sec \theta \tan \theta d\theta }{a\sqrt{{\sec }^2\theta -1}}$

$\Rightarrow \int \frac{\sec \theta \tan \theta d\theta }{\sqrt{{\sec }^2\theta -1}}$

$\Rightarrow \int \frac{\sec \theta \tan \theta d\theta }{\tan \theta }({\sec }^2\theta -1={\tan }^2\theta )$

$\Rightarrow \int \sec \theta d\theta$

$\Rightarrow \log (\sec \theta +\tan \theta )+C$

Now, from

$=\log (\frac{x}{a}+\frac{\sqrt{x^2-a^2}}{a})+C$

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

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

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

Where, $\frac{1}{a}$ term can be factored from both of these

Hence, this is the answer.