Question

Evaluate: $\frac{dx}{x\sqrt{x^4-1}}$

Answer 1

$\int \frac{dx}{x\sqrt{x^4-1}}=\int \frac{xdx}{x^2\sqrt{x^4-1}}$

Substituting $x^2=\sec \theta$ where $\theta \in (0,\frac{\pi }{2})$,

$\Rightarrow 2xdx=\sec \theta \tan \theta d\theta$

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

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

$=\frac{1}{2}\int \frac{\tan \theta }{\left|\tan \theta \right|}d\theta$

$=\frac{1}{2}\int \frac{\tan \theta }{\tan \theta }d\theta =\frac{1}{2}\int 1\cdot d\theta$

$[\because \theta \in (0,\frac{\pi }{2})]$

$=\frac{\theta }{2}+c$

$=\frac{1}{2}{\sec }^{-1}{x}^2+c$ $[\because x^2=\sec \theta ]$
where $c$ is constant of integration
Hence, the required integration is $\frac{1}{2}{\sec }^{-1}{x}^2+c$