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

Let $I = \int {\frac{{dx}}{{x\sqrt {{x^4} - 1} }}}$

Let's put ${x^2} = \sec \theta \Rightarrow \theta = {\sec ^{ - 1}}{x^2}$

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

therefore,$I = \frac{1}{2}\int {\frac{{\sec \theta \cdot \tan \theta }}{{\sec \theta \tan \theta }}} d\theta = \frac{1}{2}\int d \theta = \frac{1}{2}\theta + C$

$= \frac{1}{2}{\sec ^{ - 1}}\left( {{x^2}} \right) + C$

Evaluate the following as limit of sums :