We are tasked with integrating the following expression:

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

Steps to solve:

1. Substitute: The denominator can be simplified by substituting $u = 3x^2 + 4x$.

   $du = (6x + 4) \, dx$

   Notice that we have $3x + 2$ in the numerator, which can be factored from $6x + 4$ as follows:

   $6x + 4 = 2(3x + 2)$

   Therefore,

   $du = 2(3x + 2) \, dx$

   Solving for $3x + 2 \, dx$:

   $(3x + 2) \, dx = \frac{du}{2}$

2. Substitute in the integral:

   Now, the integral becomes:

   $\int \frac{\frac{du}{2}}{u^3} = \frac{1}{2} \int u^{-3} \, du$

3. Integrate:

   The integral of $u^{-3}$ is:

   $\int u^{-3} \, du = \frac{u^{-2}}{-2} = -\frac{1}{2u^2}$

4. Final solution:

   Substituting back $u = 3x^2 + 4x$:

   $-\frac{1}{4(3x^2 + 4x)^2} + C$

Thus, the integral is:

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

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

Here are 5 related questions to expand your understanding:

1. How would the result change if the exponent on the denominator was different?
2. Can you use partial fractions for integrals like this?
3. What are the rules for substitution in integrals?
4. How does changing the coefficients of $x$ in the integrand affect the solution?
5. Can this method be used for trigonometric integrals?

Tip: For tricky integrals, substitution often simplifies the process by reducing complex expressions into more manageable forms.