Question

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

Answer 1

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

Substitute $x=\sec u$ , $u={\sec }^{-1}x$ . $dx=\sec u\tan udu$

$=\int \frac{\sec u\tan u}{{\sec }^2\sqrt{{\sec }^{2}u-1}}du$

$\because {\sec }^{2}u-1={\tan }^{2}u$

$=\int \frac{1}{\sec u}du$

$=\int \cos udu$

$=\sin u$

$=\sin \left({\sec }^{-1}x\right)$

$=\frac{\sqrt{x^2-1}}{x}$