To solve the integral \(\int \ln(12x) \, dx\), we can use integration by parts. The formula for integration by parts is:

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

Let's choose:
- \( u = \ln(12x) \), so \( du = \frac{1}{12x} \cdot 12 \, dx = \frac{1}{x} \, dx \)
- \( dv = dx \), so \( v = x \)

Now, apply the integration by parts formula:

\[
\int \ln(12x) \, dx = x \ln(12x) - \int x \cdot \frac{1}{x} \, dx
\]

Simplifying the integral:

\[
\int \ln(12x) \, dx = x \ln(12x) - \int 1 \, dx
\]

\[
\int \ln(12x) \, dx = x \ln(12x) - x + C
\]

Where \( C \) is the constant of integration.

So, the solution is:

\[
\int \ln(12x) \, dx = x \ln(12x) - x + C
\]

Would you like further details or have any questions about this solution?

Here are some related questions you might find helpful:
1. How do you apply integration by parts in more complex cases?
2. What are other methods to integrate logarithmic functions?
3. Can integration by parts be used for \(\int \ln(x) \, dx\)?
4. How does the choice of \( u \) and \( dv \) affect the outcome of integration by parts?
5. What are some common mistakes to avoid when using integration by parts?

**Tip:** Integration by parts is particularly useful when dealing with the product of a polynomial and a logarithmic or exponential function.

Learn how to integrate ln(12x) step-by-step with integration by parts. Understand the formula used and follow a detailed solution. Suitable for college-level students studying calculus and integration.

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