Answer
$\frac{5}{(x-1)(x+4)}=\frac{1}{x-1}+\frac{-1}{x+4}$
Work Step by Step
$\frac{5}{(x-1)(x+4)}=\frac{A}{x-1}+\frac{B}{x+4}$,
$5=A(x+4)+B(x-1)$,
$5=Ax+4A+Bx-B$,
$5=(A+B)x+(4A-B)$,
$(A+B)x=0x$,
$A+B=0$ and $4A-B=5$:
$\begin{cases}
& 4A & -B & = 5&\\
& A & +B & = 0&\\
& -- & -- & -- \\
& 5A & +0 & = 5
\end{cases}$
$A=1$, substituting back into the equation, $1+B=0, B=-1$
thus,
$\frac{5}{(x-1)(x+4)}=\frac{1}{x-1}+\frac{-1}{x+4}$
