Question

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

Answer 1

$$\begin{gathered} \int \frac{dx}{x\sqrt{x^4-1}} \\ =\int \frac{xdx}{x^2\sqrt{{\left(x^2\right)}^2-1}} \\ \mathrm{Let}{x}^2=t \\ \Rightarrow 2x=\frac{dt}{dx} \\ \Rightarrow xdx=\frac{dt}{2} \\ \mathrm{Now},\int \frac{xdx}{x^2\sqrt{{\left(x^2\right)}^2-1}} \\ =\frac{1}{2}\int \frac{dt}{t\sqrt{t^2-1}} \\ =\frac{1}{2}{\sec }^{-1}\left(t\right)+C \\ =\frac{1}{2}{\sec }^{-1}\left(x^2\right)+C \end{gathered}$$