Express $\frac{3x^2 - 4}{x^2(3x - 2)}$ in partial fractions. Hence, evaluate $\int \frac{3x^2 - 4}{x^2(3x - 2)} \, dx$.

1. State the correct form for the partial fractions

Let $\frac{3x^2 - 4}{x^2(3x - 2)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{3x - 2}$

2. Form the basic equation by clearing the denominator

$3x^2 - 4 = Ax(3x - 2) + B(3x - 2) + Cx^2$

3. Find the values of $B$ and $C$ by substituting suitable values of $x$

Substitute $x = 0$:
$-4 = B(-2) \implies B = 2$

Substitute $x = \frac{2}{3}$:
$3\left(\frac{2}{3}\right)^2 - 4 = C\left(\frac{2}{3}\right)^2$
$\frac{4}{3} - 4 = \frac{4}{9}C$
$-\frac{8}{3} = \frac{4}{9}C \implies C = -6$

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

Compare coefficients of $x^2$:
$3 = 3A + C$
$3 = 3A - 6$
$3A = 9 \implies A = 3$

1. Express the integrand using the partial fractions found in part (a)

From part (a), we first find the partial fractions for $\frac{3x^2 - 4}{x^2(3x - 2)}$.
Let $\frac{3x^2 - 4}{x^2(3x - 2)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{3x - 2}$.
$3x^2 - 4 = Ax(3x - 2) + B(3x - 2) + Cx^2$
Substituting $x = 0$ gives $-4 = -2B \implies B = 2$.
Substituting $x = \frac{2}{3}$ gives $3(\frac{4}{9}) - 4 = C(\frac{4}{9}) \implies -\frac{8}{3} = \frac{4}{9}C \implies C = -6$.
Comparing coefficients of $x^2$ gives $3 = 3A + C \implies 3 = 3A - 6 \implies A = 3$.
Thus, $\frac{3x^2 - 4}{x^2(3x - 2)} = \frac{3}{x} + \frac{2}{x^2} - \frac{6}{3x - 2}$.

$\int \frac{3x^2 - 4}{x^2(3x - 2)} \, dx = \int \left( \frac{3}{x} + 2x^{-2} - \frac{6}{3x - 2} \right) \, dx$

2. Integrate each term using standard integration rules

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