Answer
$\frac{3}{2 (4x-3)} - \frac{1}{2(2x-1)}$
Work Step by Step
$\frac{x}{(8x^2 - 10x + 3)}$ = $\frac{A}{4x-3} + \frac{B}{2x-1}$
Multiply both sides by (4x-3) (2x-1)
$x = A(2x-1) + B (4x-3)$
$2A + 4B = 1$
$-A -3B = 0$
Thus A = 3/2 and B = -1/2
$\frac{3}{2 (4x-3)} - \frac{1}{2(2x-1)}$
