Question

Evaluate :
$\int \frac{dx}{x\sqrt{x^2-4}}$

Answer 1

To Evaluate

$\int \frac{dx}{x\sqrt{x^2-4}}$

Let , $u=\sqrt{x^2-4}$

$u^2=x^2-4$

So that $2udu=2xdx$

$udu=xdx$

On Apply "u -substitution" we have

$=\int \frac{du}{u^2+4}$

$=\int \frac{du}{u^2+2^2}$

On Integrating this we have

$=\frac{1}{2}ta{n}^{-1}\left[\frac{u}{2}\right]+C$

$=\frac{1}{2}ta{n}^{-1}\left[\frac{\sqrt{x^2-4}}{2}\right]+C$

Where C is the Integration Constant