Let $I=\int \frac{1}{x^4-1}dx$ Let $\frac{1}{x^4-1}=\frac{1}{(x+1)(x-1)(x^2+1)}$ $=\frac{A}{x+1}+\frac{B}{x-1}+\frac{Cx+D}{x^2+1}$ Putting x = –1 in (i), we get 1 = A(–1 – 1)(1 + 1) ⇒ 1 = A(– 4) ⇒ A $=-\frac{1}{4}$Putting x = 1 in (i), we get 1 = B (1 + 1) (1 + 1) ⇒ 1 = B(2)(2) ⇒ B $=\frac{1}{4}$ Comparing coefficients of $x^3$ and constant in (i) on both sides, we get A + B + C = 0 and – A + B – D = 1 $\Rightarrow -\frac{1}{4}+\frac{1}{4}+C=0\Rightarrow C=0$ and $\frac{1}{4}+\frac{1}{4}-D=1\Rightarrow D=-\frac{1}{2}$$\therefore \frac{1}{x^4-1}$ $=\frac{1}{4(x+1)}+\frac{1}{4(x-1)}+\frac{-1}{2(x^2+1)}$$\therefore \int \frac{1}{x^4-1} dx$ $=-\frac{1}{4}\int \frac{dx}{x+1}+\frac{1}{4}\int \frac{dx}{x-1}-\frac{1}{2}\int \frac{dx}{x^2+1}$$=-\frac{1}{4}\log |x+1|+\frac{1}{4}\log |x-1|-\frac{1}{2}\tan^{-1}x+C$$=\frac{1}{4}\log \left|\frac{x-1}{x+1}\right|-\frac{1}{2}\tan^{-1}x+C$