We are tasked with integrating the expression:

$\int \frac{3(3x + 2y)}{x(x + y)} \, dx$

Step 1: Simplify the expression

We start by simplifying the integrand:

$\frac{3(3x + 2y)}{x(x + y)} = 3 \cdot \frac{3x + 2y}{x(x + y)}$

Now, break up the numerator $3x + 2y$:

$3 \cdot \left( \frac{3x}{x(x + y)} + \frac{2y}{x(x + y)} \right) = 3 \left( \frac{3}{x + y} + \frac{2y}{x(x + y)} \right)$

Step 2: Separate the integrals

Now, we can separate the integrals for each term:

$3 \int \left( \frac{3}{x + y} + \frac{2y}{x(x + y)} \right) \, dx$

This gives us two integrals to solve:

$3 \int \frac{3}{x + y} \, dx + 3 \int \frac{2y}{x(x + y)} \, dx$

Step 3: Solve the first integral

The first integral is straightforward:

$3 \int \frac{3}{x + y} \, dx$

Since $y$ is treated as a constant with respect to $x$, let $u = x + y$, which gives $du = dx$. This transforms the integral into:

$3 \cdot 3 \int \frac{1}{u} \, du = 9 \ln |x + y|$

Step 4: Solve the second integral

The second integral is:

$3 \int \frac{2y}{x(x + y)} \, dx$

Factor out constants:

$6y \int \frac{1}{x(x + y)} \, dx$

Using partial fractions decomposition:

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

Solving for $A$ and $B$, we get:

$A = \frac{1}{y}, \quad B = -\frac{1}{y}$

Thus:

$\frac{1}{x(x + y)} = \frac{1}{y} \left( \frac{1}{x} - \frac{1}{x + y} \right)$

Substituting this into the integral:

$6y \cdot \frac{1}{y} \int \left( \frac{1}{x} - \frac{1}{x + y} \right) \, dx = 6 \left( \ln |x| - \ln |x + y| \right)$

Step 5: Combine results

Now, combining both integrals, we get:

$9 \ln |x + y| + 6 \left( \ln |x| - \ln |x + y| \right)$

Simplifying this expression:

$3 \ln |x + y| + 6 \ln |x|$

Final Answer:

$\int \frac{3(3x + 2y)}{x(x + y)} \, dx = 3 \ln |x + y| + 6 \ln |x| + C$

Would you like more details or have any questions?

Here are 5 related questions to expand this topic:

1. How do we handle integration when both variables $x$ and $y$ are present?
2. What is the method of partial fraction decomposition used in integration?
3. How does substitution simplify integrals with linear terms?
4. Can you explain the importance of the constant of integration?
5. How does the logarithmic function behave in integrals like this?

Tip: Always check if variables other than the one of integration are constants before applying simplification techniques.