Integrate the rational function: $\frac{1}{x^{4}-1}$

(N/A) We have $\frac{1}{x^{4}-1} = \frac{1}{(x^{2}-1)(x^{2}+1)} = \frac{1}{(x+1)(x-1)(x^{2}+1)}$.
Using partial fractions,let $\frac{1}{(x+1)(x-1)(x^{2}+1)} = \frac{A}{x+1} + \frac{B}{x-1} + \frac{Cx+D}{x^{2}+1}$.
Multiplying by $(x+1)(x-1)(x^{2}+1)$,we get $1 = A(x-1)(x^{2}+1) + B(x+1)(x^{2}+1) + (Cx+D)(x^{2}-1)$.
Comparing coefficients of $x^{3}, x^{2}, x,$ and the constant term:
$A+B+C = 0$
$-A+B+D = 0$
$A+B-C = 0$
$-A+B-D = 1$
Solving these equations,we get $A = -\frac{1}{4}, B = \frac{1}{4}, C = 0, D = -\frac{1}{2}$.
Thus,$\frac{1}{x^{4}-1} = -\frac{1}{4(x+1)} + \frac{1}{4(x-1)} - \frac{1}{2(x^{2}+1)}$.
Integrating both sides,$\int \frac{1}{x^{4}-1} dx = -\frac{1}{4} \int \frac{1}{x+1} dx + \frac{1}{4} \int \frac{1}{x-1} dx - \frac{1}{2} \int \frac{1}{x^{2}+1} dx$.
$= -\frac{1}{4} \log |x+1| + \frac{1}{4} \log |x-1| - \frac{1}{2} \tan^{-1} x + C$.
$= \frac{1}{4} \log \left| \frac{x-1}{x+1} \right| - \frac{1}{2} \tan^{-1} x + C$,where $C$ is the constant of integration.