To evaluate the integral of $\ln(3x)$ from $x = 1$ to $x = 9$, we need to integrate the function $\ln(3x)$ with respect to $x$.

Step-by-step Solution:

1. Integral Setup: $\int \ln(3x) \, dx$

2. Integration by Parts: Let: $u = \ln(3x) \implies du = \frac{1}{3x} \cdot 3 \, dx = \frac{dx}{x}$ $dv = dx \implies v = x$

   Using the integration by parts formula, $\int u \, dv = uv - \int v \, du$, we get:

   $\int \ln(3x) \, dx = x \ln(3x) - \int x \cdot \frac{dx}{x}$ $= x \ln(3x) - \int dx$ $= x \ln(3x) - x + C$

3. Definite Integral from $x = 1$ to $x = 9$: Now we substitute the limits $x = 1$ and $x = 9$:

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

   Calculating at $x = 9$:

   $9 \ln(3 \times 9) - 9 = 9 \ln(27) - 9$

   Since $\ln(27) = \ln(3^3) = 3 \ln(3)$, this becomes:

   $9 \cdot 3 \ln(3) - 9 = 27 \ln(3) - 9$

   Calculating at $x = 1$:

   $1 \ln(3 \times 1) - 1 = \ln(3) - 1$

4. Compute the Result:

   $\text{Integral} = (27 \ln(3) - 9) - (\ln(3) - 1)$ $= 27 \ln(3) - 9 - \ln(3) + 1$ $= 26 \ln(3) - 8$

Final Answer:

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

Would you like more details on any step, or do you have any questions?

Related Questions:

1. How do you perform integration by parts for other functions?
2. What are the rules for logarithm properties and how do they apply in integration?
3. How would the solution change if the integrand were $\ln(ax)$ for any constant $a$?
4. What are some applications of definite integrals in real-world problems?
5. Can we use substitution methods for logarithmic integrals, and if so, how?

Tip:

When integrating logarithmic functions, remember to consider integration by parts as it is often the most effective technique.