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

\[ = \frac{1}{2}{\sec ^{ – 1}}\left( {{x^2}} \right) + C\]

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