To solve the integral $\int \ln(9x + 6) \, dx$ using integration by parts, we follow these steps:

Step 1: Set up for Integration by Parts

The formula for integration by parts is:

$\int u \, dv = uv - \int v \, du$

We need to choose $u$ and $dv$ appropriately.

• Let $u = \ln(9x + 6)$. Then, $du = \frac{9}{9x + 6} \, dx$.
• Let $dv = dx$. Then, $v = x$.

Step 2: Apply the Integration by Parts Formula

Using the integration by parts formula:

$\int \ln(9x + 6) \, dx = x \ln(9x + 6) - \int x \left(\frac{9}{9x + 6}\right) \, dx$

Step 3: Simplify the Remaining Integral

We need to simplify $\int x \left( \frac{9}{9x + 6} \right) \, dx$. Let’s rewrite the expression:

$\int x \left( \frac{9}{9x + 6} \right) \, dx = 9 \int \frac{x}{9x + 6} \, dx$

Use substitution: Let $u = 9x + 6$, so $du = 9 \, dx$, and therefore $dx = \frac{du}{9}$. Also, $x = \frac{u - 6}{9}$.

Substitute into the integral:

$9 \int \frac{x}{9x + 6} \, dx = 9 \int \frac{\frac{u - 6}{9}}{u} \cdot \frac{du}{9}$

This simplifies to:

$\int \frac{u - 6}{u} \, du = \int \left( 1 - \frac{6}{u} \right) \, du$

Step 4: Solve the Simplified Integral

Now we can integrate the two terms:

$\int 1 \, du - \int \frac{6}{u} \, du = u - 6 \ln |u| + C$

Substitute back $u = 9x + 6$:

$u - 6 \ln |u| + C = (9x + 6) - 6 \ln |9x + 6| + C$

Step 5: Final Answer

Now substitute this result back into the integration by parts expression:

$\int \ln(9x + 6) \, dx = x \ln(9x + 6) - \left[ (9x + 6) - 6 \ln |9x + 6| \right] + C$

Simplifying:

$\int \ln(9x + 6) \, dx = x \ln(9x + 6) - (9x + 6) + 6 \ln(9x + 6) + C$

This is the evaluated integral.

Would you like further clarifications or details on any of the steps?

Here are some related questions:

1. How can substitution simplify complex integrals?
2. What is the general strategy for choosing $u$ and $dv$ in integration by parts?
3. How can you check the correctness of an integral using differentiation?
4. What are other techniques for integrating logarithmic functions?
5. How does integration by parts handle products of functions?

Tip: When applying integration by parts, it's important to choose $u$ and $dv$ carefully so that the remaining integral becomes simpler to evaluate.