If ∫ d x x x 4 - 1 = 1 k sec - 1 x k , then the value of k =

 Given that

$\int \frac{dx}{x\left(\sqrt{x^4-1}\right)}$

Put $x^2 = \sec \theta$

$\Rightarrow 2x dx = \sec \theta \tan \theta d\theta$

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

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

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

$=\frac{1}{2}{\sec }^{-1}{x}^2 + c$

Given that it's equal to $\frac{1}{k}{\sec }^{-1}\left(x^k\right)$

By comparing $k =2.$