Answer
$\frac{3}{x} + \frac{2}{2x-1}$
Work Step by Step
$\frac{8x-3}{(2x^2 - x)}$ = $\frac{A}{x} + \frac{B}{2x-1}$
Multiply both sides by (x) (2x-1)
$8x-3 = A(2x-1) + B (x)$
$2A + B = 8$
$-A = -3$
Thus A = 3 and B = 2
$\frac{3}{x} + \frac{2}{2x-1}$
