Evaluate the integral ln3x from x=1 to x=9

To evaluate the integral of \(\ln(3x)\) from \(x = 1\) to \(x = 9\), we first need to find the antiderivative of \(\ln(3x)\).

### Step 1: Set Up the Integral
The integral we need to evaluate is:

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

### Step 2: Use Integration by Parts
To integrate \(\ln(3x)\), we use integration by parts. Recall the formula for integration by parts:

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

Let's choose:
- \(u = \ln(3x)\) \(\implies du = \frac{1}{3x} \cdot 3 \, dx = \frac{1}{x} \, dx\)
- \(dv = dx\) \(\implies v = x\)

Applying the integration by parts formula:

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

Simplify the integral:

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

### Step 3: Apply the Limits
Now, apply the limits from \(x = 1\) to \(x = 9\):

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

Evaluate this expression at \(x = 9\) and \(x = 1\):

1. At \(x = 9\):
   \[
   9 \ln(27) - 9 = 9 \ln(3^3) - 9 = 9 \cdot 3 \ln(3) - 9 = 27 \ln(3) - 9
   \]

2. At \(x = 1\):
   \[
   1 \ln(3) - 1 = \ln(3) - 1
   \]

### Step 4: Subtract the Results
Subtract the value at \(x = 1\) from the value at \(x = 9\):

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

Combine like terms:

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

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

Would you like more details or have any questions? Here are some related topics:

1. How to perform integration by parts for other functions.
2. Understanding natural logarithms in integration.
3. Techniques for solving definite integrals.
4. The properties of logarithmic functions.
5. Applications of integration by parts in physics.

**Tip:** When using integration by parts, always choose \(u\) and \(dv\) such that the resulting integral is simpler than the original.

Learn how to evaluate the integral of ln(3x) from x=1 to x=9 with a step-by-step solution using integration by parts. This problem covers advanced calculus concepts including integration techniques and natural logarithms. Suitable for advanced high school students and above.

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