Why did you multiply 1/3x by 3dx instead of dx

I see where the confusion might come from. When I took the derivative \(du = \frac{1}{3x} \cdot 3 \, dx\), it was indeed confusingly written. Let's clarify the steps to avoid confusion. 

### Correct Differentiation:

When we perform integration by parts for \( \int \ln(3x) \, dx \), we choose:

- \( u = \ln(3x) \)
- \( dv = dx \)

Now, to find \(du\), we need to differentiate \( u = \ln(3x) \):

\[
u = \ln(3x)
\]

Applying the chain rule:

\[
\frac{d}{dx}[\ln(3x)] = \frac{1}{3x} \cdot 3 = \frac{1}{x}
\]

Thus,

\[
du = \frac{1}{x} \, dx
\]

There was no need to multiply by 3 after differentiating. It was directly \(\frac{1}{x} \, dx\). Here’s the corrected version:

### Integration by Parts Step-by-Step:

Given:
\[
\int \ln(3x) \, dx
\]

1. **Choose \(u\) and \(dv\):**

   \[
   u = \ln(3x), \quad dv = dx
   \]

2. **Compute \(du\) and \(v\):**

   \[
   du = \frac{1}{x} \, dx, \quad v = x
   \]

3. **Apply Integration by Parts Formula:**

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

   Substituting the expressions:

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

4. **Evaluate the Definite Integral:**

   From \(x = 1\) to \(x = 9\):

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

   At \(x = 9\):

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

   At \(x = 1\):

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

5. **Final Answer:**

   \[
   (27 \ln(3) - 9) - (\ln(3) - 1) = 26 \ln(3) - 8
   \]

So, the integral is:

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

Thank you for catching that point! If you have further questions or need more clarifications, feel free to ask!

### Related Questions:

1. What is the chain rule in calculus, and how is it applied in differentiation?
2. Can you provide more examples of using integration by parts?
3. How do we approach integrals that involve products of logarithmic and polynomial functions?
4. What are common mistakes to avoid when performing integration by parts?
5. How does the choice of \(u\) and \(dv\) affect the complexity of an integral?

### Tip:
When differentiating composite functions, always apply the chain rule carefully, and clearly identify each step to avoid errors.

Learn the correct application of the chain rule in differentiation and integration by parts with a detailed explanation. This problem focuses on the integration of ln(3x) and clarifies steps to avoid confusion. Suitable for college-level students studying calculus.

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