Question

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

Answer 1

$\frac{1}{(x^4-1)}=\frac{1}{(x^2-1)(x^2+1)}=\frac{1}{(x+1)(x-1)(1+x^2)}$
Let $\frac{1}{(x+1)(x-1)(1+x^2)}=\frac{A}{(x+1)}+\frac{B}{(x-1)}+\frac{Cx+D}{(x^2+1)}$
$\Rightarrow 1=A(x-1)(x^2+1)+B(x+1)(x^2+1)+(Cx+D)(x^2+1)$
$\Rightarrow 1=A(x^3+x-x^2-1)+B(x^3+x+x^2+1)+C{x}^3+D{x}^2-Cx-D$
$\Rightarrow 1=(A+B+C)x^3+(-A+B+D)x^2+(A+B-C)x+(-A+B-D)$
Equating the coefficient of $x^3,x^2,x$, and constant term, we obtain
$A+B+C=0$
$-A+B+D=0$
$A+B-C=0$
$-A+B-D=1$
On solving these equations, we obtain
$A=-\frac{1}{4},B=\frac{1}{4},C=0$, and $D=-\frac{1}{2}$
$\therefore \frac{1}{x^4-1}=\frac{-1}{4(x+1)}+\frac{1}{4(x-1)}-\frac{1}{2(x^2+1)}$
$\Rightarrow \int \frac{1}{x^4-1}dx=-\frac{1}{4}\log |x-1|+\frac{1}{4}\log |x-1|-\frac{1}{2}{\tan }^{-1}x+C$
$=\frac{1}{4}\log \left|\frac{x-1}{x+1}\right|-\frac{1}{2}{\tan }^{-1}x+C$