Question

Integrate the following function $w.r.t.x$ :
$\frac{1}{x^2\sqrt{x^2-1}}$

Answer 1

$\int \frac{dx}{x^2\sqrt{x^2-1}}$

Let $x\sec \theta \Rightarrow dx=\sec \theta \tan \theta d\theta$

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

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

$\int \cos \theta d\theta$

$=\sin \theta +c$ where $c$ is the constant of integration.

We have $x=\sec \theta \Rightarrow \frac{1}{x}=\cos \theta$

$\Rightarrow \sin \theta =\sqrt{1-{\left(\frac{1}{x}\right)}^2}=\frac{\sqrt{x^2-1}}{x}$

$\therefore \int \frac{dx}{x^2\sqrt{x^2-1}}=\frac{\sqrt{x^2-1}}{x}+c$