Answer
$\frac{2x+1}{(x+2)(x-1)}=\frac{1}{x+2}+\frac{1}{x-1}$
Work Step by Step
$\frac{2x+1}{(x^2+x-2)}$, factorising $x^2+x-2=(x+2)(x-1)$:
$\frac{2x+1}{(x+2)(x-1)}=\frac{A}{x+2}+\frac{B}{x-1}$,
$2x+1=A(x-1)+B(x+2)$,
$2x+1=Ax-A+Bx+2B$,
$2x+1=(A+B)x+(-A+2B)$,
$(A+B)x=2x$,
$A+B=2$ and $-A+2B=1$:
$\begin{cases}
& A & +B & = 2&\\
& -A & +2B & = 1&\\
& -- & -- & -- \\
& 0 & +3B & = 3
\end{cases}$
$B=1$, substituting back into the equation, $A+1=2, A=1$
thus,
$\frac{2x+1}{(x+2)(x-1)}=\frac{1}{x+2}+\frac{1}{x-1}$
