Question

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

Answer 1

$\begin{array}{c}{\int }^{}\frac{1}{x^4-1}dx={\int }^{}\frac{1}{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}dx\\ ={\int }^{}\left(-\frac{1}{2\left(x^2+1\right)}-\frac{1}{4\left(x+1\right)}+\frac{1}{4\left(x-1\right)}\right)dx\\ =-\frac{\ln \left(x+1\right)}{4}-\frac{{\tan }^{-1}\left(x\right)}{2}+\frac{\ln \left(x-1\right)}{4}+C\end{array}$