[Cambridge/9233/2004] Use partial fractions to evaluate $\int_{2}^{3} \frac{9x^2}{(x - 1)^2 (x + 2)} \, dx$, giving your answer in an exact form.

1. Express the integrand in terms of partial fractions.

Let $\frac{9x^2}{(x - 1)^2 (x + 2)} = \frac{A}{x - 1} + \frac{B}{(x - 1)^2} + \frac{C}{x + 2}$

Multiply throughout by $(x - 1)^2 (x + 2)$:
$9x^2 = A(x - 1)(x + 2) + B(x + 2) + C(x - 1)^2$

2. Solve for the unknown constants $A$, $B$, and $C$.

Substitute $x = 1$:
$9(1)^2 = B(1 + 2) \implies 3B = 9 \implies B = 3$

Substitute $x = -2$:
$9(-2)^2 = C(-2 - 1)^2 \implies 9C = 36 \implies C = 4$

Substitute $x = 0$:
$9(0)^2 = A(0 - 1)(0 + 2) + B(0 + 2) + C(0 - 1)^2$
$0 = -2A + 2(3) + 4(-1)^2$
$0 = -2A + 6 + 4 \implies 2A = 10 \implies A = 5$

Thus, the partial fraction decomposition is:
$\frac{9x^2}{(x - 1)^2 (x + 2)} = \frac{5}{x - 1} + \frac{3}{(x - 1)^2} + \frac{4}{x + 2}$

3. Integrate the decomposed partial fractions.

$\int \left( \frac{5}{x - 1} + \frac{3}{(x - 1)^2} + \frac{4}{x + 2} \right) \, dx$

$= 5 \ln |x - 1| - \frac{3}{x - 1} + 4 \ln |x + 2|$

4. Evaluate the definite integral by substituting the limits.

Substitute the limits from $2$ to $3$:
$= \left[ 5 \ln |x - 1| - \frac{3}{x - 1} + 4 \ln |x + 2| \right]_{2}^{3}$

Upper limit ($x = 3$):
$5 \ln(2) - \frac{3}{2} + 4 \ln(5)$

Lower limit ($x = 2$):
$5 \ln(1) - 3 + 4 \ln(4) = 0 - 3 + 4(2 \ln 2) = -3 + 8 \ln 2$

Subtract the lower limit evaluation from the upper limit evaluation:
$= \left( 5 \ln 2 - \frac{3}{2} + 4 \ln 5 \right) - \left( -3 + 8 \ln 2 \right)$
$= \frac{3}{2} - 3 \ln 2 + 4 \ln 5$