To integrate the function \( \ln(3x) \) from \( x = 1 \) to \( x = 9 \), we need to solve the definite integral:

\[
\int_{1}^{9} \ln(3x) \, dx
\]

This integral can be evaluated using integration by parts. Recall the formula for integration by parts:

\[
\int u \, dv = uv - \int v \, du
\]

Here, let's set:
- \( u = \ln(3x) \) (so that \( du = \frac{1}{x} \, dx \))
- \( dv = dx \) (so that \( v = x \))

Now, applying integration by parts:

\[
\int \ln(3x) \, dx = x \ln(3x) - \int x \cdot \frac{1}{x} \, dx
\]

The integral simplifies to:

\[
\int \ln(3x) \, dx = x \ln(3x) - \int dx
\]

\[
\int \ln(3x) \, dx = x \ln(3x) - x + C
\]

Now, we evaluate this from \( x = 1 \) to \( x = 9 \):

\[
\left[ x \ln(3x) - x \right]_{1}^{9} = \left[ 9 \ln(27) - 9 \right] - \left[ 1 \ln(3) - 1 \right]
\]

Substitute and simplify:

\[
= 9 \ln(27) - 9 - (1 \ln(3) - 1)
\]

Since \( \ln(27) = 3\ln(3) \), we get:

\[
= 9(3\ln(3)) - 9 - (\ln(3) - 1)
\]

\[
= 27\ln(3) - 9 - \ln(3) + 1
\]

\[
= 26\ln(3) - 8
\]

Thus, the value of the definite integral is:

\[
\int_{1}^{9} \ln(3x) \, dx = 26\ln(3) - 8
\]

Would you like more details or have any questions?

Here are some related questions you might find interesting:
1. How do you apply the method of integration by parts in general?
2. What is the interpretation of the natural logarithm function in calculus?
3. How do you integrate functions involving natural logarithms?
4. How can you derive the formula for integration by parts?
5. How do definite integrals differ from indefinite integrals in interpretation?

**Tip:** When integrating functions involving logarithms, integration by parts is often a useful method.

Explore the solution to integrate ln(3x) from x=1 to x=9 using integration by parts method. This calculus problem covers advanced concepts in integration and natural logarithms.

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