Answer
$\frac{4}{(x-2)(x+2)}=\frac{1}{x-2}+\frac{-1}{x+2}$
Work Step by Step
$\frac{4}{(x^2-4)}$, factorising $x^2-4=(x-2)(x+2)$:
$\frac{4}{(x-2)(x+2)}=\frac{A}{x-2}+\frac{B}{x+2}$,
$4=A(x+2)+B(x-2)$,
$4=Ax+2A+Bx-2B$,
$4=(A+B)x+(2A-2B)$,
$(A+B)x=0x$,
$A+B=0$ and $2A-2B=4, A-B=2$:
$\begin{cases}
& A & -B & = 2&\\
& A & +B & = 0&\\
& -- & -- & -- \\
& 2A & +0 & = 2
\end{cases}$
$A=1$, substituting back into the equation, $1+B=0, B=-1$
thus,
$\frac{4}{(x-2)(x+2)}=\frac{1}{x-2}+\frac{-1}{x+2}$
