The problem asks us to evaluate the integral of the function $\frac{1}{(x+3)(x+6)}$.

1. Problem Description

The problem asks us to evaluate the integral of the function $\frac{1}{(x+3)(x+6)}$.

2. Solution Steps

We will use partial fraction decomposition to solve the integral. We want to express $\frac{1}{(x+3)(x+6)}$ in the form $\frac{A}{x+3} + \frac{B}{x+6}$.

$$\frac{1}{(x+3)(x+6)} = \frac{A}{x+3} + \frac{B}{x+6}$$

Multiply both sides by $(x+3)(x+6)$ to get:

$$1 = A(x+6) + B(x+3)$$

To solve for $A$ and $B$, we can use the following system of equations by choosing appropriate values for $x$.

Let $x = -3$. Then

$1 = A(-3+6) + B(-3+3)$

$1 = 3A + 0$

$A = \frac{1}{3}$

Let $x = -6$. Then

$1 = A(-6+6) + B(-6+3)$

$1 = 0 - 3B$

$B = -\frac{1}{3}$

So, we have

$$\frac{1}{(x+3)(x+6)} = \frac{1/3}{x+3} - \frac{1/3}{x+6}$$

Now we integrate:

$$\int \frac{1}{(x+3)(x+6)} dx = \int \left( \frac{1/3}{x+3} - \frac{1/3}{x+6} \right) dx$$

$$= \frac{1}{3} \int \frac{1}{x+3} dx - \frac{1}{3} \int \frac{1}{x+6} dx$$

$$= \frac{1}{3} \ln |x+3| - \frac{1}{3} \ln |x+6| + C$$

We can combine the logarithms:

$$= \frac{1}{3} (\ln |x+3| - \ln |x+6|) + C$$

$$= \frac{1}{3} \ln \left| \frac{x+3}{x+6} \right| + C$$

3. Final Answer

The integral is $\frac{1}{3} \ln \left| \frac{x+3}{x+6} \right| + C$.