Express $\frac{x + 7}{(x^2 - 9)(x - 3)}$ as the sum of 3 partial fractions.

1. Factorise the denominator completely

The denominator is $(x^2 - 9)(x - 3)$.
Since $x^2 - 9 = (x - 3)(x + 3)$,
$(x^2 - 9)(x - 3) = (x - 3)(x + 3)(x - 3) = (x - 3)^2(x + 3)$.

2. Set up the partial fraction decomposition

Let $\frac{x + 7}{(x - 3)^2(x + 3)} = \frac{A}{x - 3} + \frac{B}{(x - 3)^2} + \frac{C}{x + 3}$.
Multiplying throughout by $(x - 3)^2(x + 3)$ gives:
$x + 7 = A(x - 3)(x + 3) + B(x + 3) + C(x - 3)^2$

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

Substitute $x = 3$:
$3 + 7 = A(0) + B(3 + 3) + C(0)$
$10 = 6B$
$B = \frac{5}{3}$

4. Find the value of $C$ by substituting $x = -3$

Substitute $x = -3$:
$-3 + 7 = A(0) + B(0) + C(-3 - 3)^2$
$4 = 36C$
$C = \frac{1}{9}$

5. Find the value of $A$ by substituting $x = 0$

Substitute $x = 0$:
$0 + 7 = A(0 - 3)(0 + 3) + B(0 + 3) + C(0 - 3)^2$
$7 = -9A + 3B + 9C$
$7 = -9A + 3\left(\frac{5}{3}\right) + 9\left(\frac{1}{9}\right)$
$7 = -9A + 5 + 1$
$9A = -1$
$A = -\frac{1}{9}$

6. Write the final partial fraction expression

$\frac{x + 7}{(x^2 - 9)(x - 3)} = -\frac{1}{9(x - 3)} + \frac{5}{3(x - 3)^2} + \frac{1}{9(x + 3)}$