Answer
$\displaystyle \frac{3x}{(x+2)(x-1)}=\frac{2}{x+2}+\frac{1}{x-1} $
Work Step by Step
This is Case 1: Q has only nonrepeated linear factors.
$\displaystyle \frac{3x}{(x+2)(x-1)}=\frac{A}{x+2}+\frac{B}{x-1} $
... write the RHs with a common denominator
$ \displaystyle \frac{3x}{(x+2)(x-1)}=\frac{A(x-1)+B(x+2)}{(x+2)(x-1)}$
... equate the numerators
$3x=Ax-A+Bx+2B$
$3x=(A+B)x+ (-A+2B)$
... equate the coefficients of the polynomials on the LHS and RHS:
$\left\{\begin{array}{llll}
3 & =A & +B & \\
0 & =-A & +2B &
\end{array}\right.$
Add the two equations,
$3=3B$
$B=1$
Back-substitute into the second equation
$0=-A+2(1)$
$A=2$
$\displaystyle \frac{3x}{(x+2)(x-1)}=\frac{2}{x+2}+\frac{1}{x-1} $
