Express $\frac{13x - 6}{x^2(2x - 3)}$ in partial fractions.

1. State the correct form for the partial fractions

Let $\frac{13x - 6}{x^2(2x - 3)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{2x - 3}$

2. Form the basic algebraic identity

$13x - 6 = Ax(2x - 3) + B(2x - 3) + Cx^2$

3. Find the value of $B$ by substituting $x = 0$

Substitute $x = 0$:
$-6 = B(-3)$
$B = 2$

4. Find the value of $C$ by substituting $x = \frac{3}{2}$

Substitute $x = \frac{3}{2}$:
$13\left(\frac{3}{2}\right) - 6 = C\left(\frac{3}{2}\right)^2$
$\frac{39}{2} - \frac{12}{2} = \frac{9}{4}C$
$\frac{27}{2} = \frac{9}{4}C$
$C = \frac{27}{2} \times \frac{4}{9} = 6$

5. Find the value of $A$ by comparing coefficients

Compare coefficients of $x^2$:
$0 = 2A + C$
$0 = 2A + 6$
$2A = -6$
$A = -3$

Therefore, $\frac{13x - 6}{x^2(2x - 3)} = -\frac{3}{x} + \frac{2}{x^2} + \frac{6}{2x - 3}$