The integral you are looking to evaluate is:

$\int \frac{4x + 10}{(3x + 2)(x - 3)} \, dx$

This type of integral can often be solved using partial fraction decomposition. The first step is to decompose the rational expression into simpler fractions.

Let's break it down:

$\frac{4x + 10}{(3x + 2)(x - 3)} = \frac{A}{3x + 2} + \frac{B}{x - 3}$

Multiply both sides by the denominator $(3x + 2)(x - 3)$ to eliminate the fractions:

$4x + 10 = A(x - 3) + B(3x + 2)$

Next, we expand both sides:

$4x + 10 = A(x - 3) + B(3x + 2) = A(x) - 3A + B(3x) + 2B$

Now, collect like terms:

$4x + 10 = (A + 3B)x + (-3A + 2B)$

For this equation to be true for all values of $x$, the coefficients of $x$ and the constant terms must be equal on both sides:

1. $A + 3B = 4$ (for the coefficient of $x$)
2. $-3A + 2B = 10$ (for the constant term)

Now solve this system of equations:

From the first equation:

$A + 3B = 4 \quad \Rightarrow \quad A = 4 - 3B$

Substitute this into the second equation:

$-3(4 - 3B) + 2B = 10$

Simplify:

$-12 + 9B + 2B = 10$

$11B = 22 \quad \Rightarrow \quad B = 2$

Substitute $B = 2$ into $A = 4 - 3B$:

$A = 4 - 3(2) = 4 - 6 = -2$

Now we substitute $A = -2$ and $B = 2$ into the partial fractions:

$\frac{4x + 10}{(3x + 2)(x - 3)} = \frac{-2}{3x + 2} + \frac{2}{x - 3}$

Now, integrate each term separately:

$\int \frac{-2}{3x + 2} \, dx + \int \frac{2}{x - 3} \, dx$

For the first integral:

$\int \frac{-2}{3x + 2} \, dx = -\frac{2}{3} \ln |3x + 2|$

For the second integral:

$\int \frac{2}{x - 3} \, dx = 2 \ln |x - 3|$

Thus, the final result is:

$-\frac{2}{3} \ln |3x + 2| + 2 \ln |x - 3| + C$

Let me know if you need further details or have any questions!

Here are some related questions for further exploration:

1. How do we handle integrals involving more complicated partial fraction decompositions?
2. What is the method for handling integrals with quadratic denominators?
3. How can we apply the technique of partial fractions to definite integrals?
4. What is the connection between logarithmic functions and rational functions in integrals?
5. How do you deal with improper fractions when performing integration?

Tip: Practice simplifying fractions and solving for the coefficients in partial fractions; this can make solving many integrals much easier!