Let's go through the steps to solve the integral:

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

Step 1: Substitution

Since the integrand involves $3x + 2$, a useful approach is to apply the substitution method. Let’s set:

$u = 3x + 2$

This simplifies the expression inside the integrand. Now, we need to find $du$ in terms of $dx$. Differentiate $u$ with respect to $x$:

$\frac{du}{dx} = 3 \quad \Rightarrow \quad dx = \frac{du}{3}$

Step 2: Rewrite the Integral

Substitute $u = 3x + 2$ and $dx = \frac{du}{3}$ into the original integral:

$\int (3x + 2)^{-4} \, dx = \int u^{-4} \cdot \frac{du}{3}$

Factor out the constant $\frac{1}{3}$:

$\frac{1}{3} \int u^{-4} \, du$

Step 3: Integrate

Now, we integrate $u^{-4}$. Recall the power rule for integration:

$\int u^n \, du = \frac{u^{n+1}}{n+1} + C \quad \text{for} \quad n \neq -1$

For $u^{-4}$, $n = -4$, so:

$\int u^{-4} \, du = \frac{u^{-3}}{-3} = -\frac{1}{3} u^{-3} + C$

Thus, the integral becomes:

$\frac{1}{3} \cdot \left(-\frac{1}{3} u^{-3} \right) + C = -\frac{1}{9} u^{-3} + C$

Step 4: Substitute Back $u = 3x + 2$

Now, substitute back $u = 3x + 2$ to express the result in terms of $x$:

$-\frac{1}{9} (3x + 2)^{-3} + C$

This is the final answer.

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Would you like more details on any step or have any questions?

Here are 5 related questions to deepen your understanding:

1. How would you handle a similar integral with a different power, such as $(3x + 2)^{-2}$?
2. What if the integrand had a coefficient in front of $x$, e.g., $5x + 4$?
3. How does the power rule for integration change when dealing with negative exponents?
4. What other techniques could be used if substitution is not straightforward?
5. How do constants affect the result of an integral?

Tip: Always check if substitution is a good approach when the integrand involves a linear expression like $ax + b$.