Express $\frac{2x^3 - 4x^2 - 5x + 3}{(x - 3)(x^2 - 9)}$ in partial fractions. Hence find $\int \frac{2x^3 - 4x^2 - 5x + 3}{(x - 3)(x^2 - 9)} \, dx$.

1. Factorise the denominator and identify that the fraction is improper.

Denominator: $(x - 3)(x^2 - 9) = (x - 3)(x - 3)(x + 3) = (x - 3)^2(x + 3)$
Expanding the denominator gives $x^3 - 3x^2 - 9x + 27$.
The degree of the numerator ($3$) is equal to the degree of the denominator ($3$), so it is an improper fraction.

2. Perform long division to extract the polynomial part.

So, $\frac{2x^3 - 4x^2 - 5x + 3}{(x - 3)(x^2 - 9)} = 2 + \frac{2x^2 + 13x - 51}{(x - 3)^2(x + 3)}$.

3. Set up the partial fraction decomposition for the proper fraction part.

Let $\frac{2x^2 + 13x - 51}{(x - 3)^2(x + 3)} = \frac{A}{x - 3} + \frac{B}{(x - 3)^2} + \frac{C}{x + 3}$.
Multiplying by $(x - 3)^2(x + 3)$ gives:
$2x^2 + 13x - 51 = A(x - 3)(x + 3) + B(x + 3) + C(x - 3)^2$

4. Solve for the constants $A$, $B$, and $C$.

Substitute $x = 3$:
$2(3)^2 + 13(3) - 51 = B(3 + 3)$
$18 + 39 - 51 = 6B \implies 6 = 6B \implies B = 1$

Substitute $x = -3$:
$2(-3)^2 + 13(-3) - 51 = C(-3 - 3)^2$
$18 - 39 - 51 = 36C \implies -72 = 36C \implies C = -2$

Compare coefficients of $x^2$:
$2 = A + C$
$2 = A - 2 \implies A = 4$

5. Write the final expression.

$\frac{2x^3 - 4x^2 - 5x + 3}{(x - 3)(x^2 - 9)} = 2 + \frac{4}{x - 3} + \frac{1}{(x - 3)^2} - \frac{2}{x + 3}$

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

From part (i), we have:
$\frac{2x^3 - 4x^2 - 5x + 3}{(x - 3)(x^2 - 9)} = 2 + \frac{4}{x - 3} + \frac{1}{(x - 3)^2} - \frac{2}{x + 3}$

Therefore, the integral becomes:
$\int \left( 2 + \frac{4}{x - 3} + \frac{1}{(x - 3)^2} - \frac{2}{x + 3} \right) \, dx$

2. Integrate each term with respect to $x$

$\int 2 \, dx + \int \frac{4}{x - 3} \, dx + \int (x - 3)^{-2} \, dx - \int \frac{2}{x + 3} \, dx$
$= 2x + 4\ln|x - 3| + \frac{(x - 3)^{-1}}{-1} - 2\ln|x + 3| + c$
$= 2x + 4\ln|x - 3| - \frac{1}{x - 3} - 2\ln|x + 3| + c$