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

1. Substitute the partial fractions obtained in part (a)

From part (a), we have:
$\frac{13x - 6}{x^2(2x - 3)} = -\frac{3}{x} + \frac{2}{x^2} + \frac{6}{2x - 3}$

$\int \frac{13x - 6}{x^2(2x - 3)} \, dx = \int \left( -\frac{3}{x} + 2x^{-2} + \frac{6}{2x - 3} \right) \, dx$

2. Integrate each term with respect to $x$

$= -3 \ln x + \frac{2x^{-1}}{-1} + \frac{6 \ln(2x - 3)}{2} + c$
$= -3 \ln x - \frac{2}{x} + 3 \ln(2x - 3) + c$