Answer
$\frac{ax+b}{(x-1)(x+1)}=\frac{a+b}{2x-2}+\frac{a-b}{2x+2}$
Work Step by Step
$\frac{ax+b}{x^2-1}$, factorising $(x^2-1)=(x-1)(x+1)$.
$\frac{ax+b}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}$,
$A(x+1)+B(x-1)$,
$Ax+A+Bx-B$,
$(A+B)x+(A-B)=ax+b$,
$\begin{cases}
A+B=a\\
A-B=b\\
-- -- -\\
2A=a+b
\end{cases}$,
thus, $A=\frac{a+b}{2}$, substituting back into the Equation, $\frac{a+b}{2}+B=a, B=\frac{a-b}{2}$,
Therefore,
$\frac{ax+b}{(x-1)(x+1)}=\frac{a+b}{2x-2}+\frac{a-b}{2x+2}$,
