Answer
$\frac{2x}{(x-1)(x+1)}=\frac{1}{x-1}+\frac{1}{x+1}$
Work Step by Step
$\frac{2x}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}$,
$2x=A(x+1)+B(x-1)$,
$2x=Ax+A+Bx-B$,
$2x=(A+B)x+(A-B)$,
$(A+B)x=2x$,
$A+B=2$ and $A-B=0, A=B$:
$A=1,B=1$
thus,
$\frac{2x}{(x-1)(x+1)}=\frac{1}{x-1}+\frac{1}{x+1}$
