Answer
$\dfrac{6}{5(x+2)}+\dfrac{8}{5(2x-1)}$
Work Step by Step
We will write the partial decomposition as:
$\dfrac{4x+2}{(x+2)(2x-1)}=\dfrac{A}{x+2}+\dfrac{B}{2x-1} ~~~~(a)$
This implies that $4x-2=A(2x-1)+B(x+2)$
Plug $x=-2$ to obtain: $A=\dfrac{6}{5}$
Now, plug $x=\dfrac{1}{2}$ to obtain: $B=\dfrac{8}{5}$
So, the equation (a) becomes:
$\dfrac{4x+2}{(x+2)(2x-1)}=\dfrac{6}{5(x+2)}+\dfrac{8}{5(2x-1)}$
