We are given the integral

$\int \frac{\ln(9x)}{x^3} \, dx$

and we need to apply integration by parts, choosing $u = \ln(9x)$. In this method, we need to find the functions $F(x)$ and $G(x)$ as indicated.

Step 1: Assign parts for integration by parts

We know that for integration by parts:

$\int u \, dv = u v - \int v \, du$

Here, we let:

• $u = \ln(9x)$, which implies $du = \frac{1}{x} \, dx$
• $dv = \frac{1}{x^3} \, dx$, so we need to find $v$.

Step 2: Integrate $dv$ to find $v$

We can integrate $dv = \frac{1}{x^3} \, dx$ as follows:

$v = \int \frac{1}{x^3} \, dx = -\frac{1}{2x^2}$

Step 3: Apply the integration by parts formula

Now, substitute into the integration by parts formula:

$\int \frac{\ln(9x)}{x^3} \, dx = \ln(9x) \left( -\frac{1}{2x^2} \right) - \int \left( -\frac{1}{2x^2} \cdot \frac{1}{x} \right) \, dx$

Simplify the integrand in the remaining integral:

$= -\frac{\ln(9x)}{2x^2} + \frac{1}{2} \int \frac{1}{x^3} \, dx$

Step 4: Complete the calculation

The second integral is:

$\int \frac{1}{x^3} \, dx = -\frac{1}{2x^2}$

Thus, the final expression is:

$-\frac{\ln(9x)}{2x^2} + \frac{1}{4x^2} + C$

Conclusion:

The function $F(x)$ is the first term obtained from the integration by parts:

$F(x) = -\frac{\ln(9x)}{2x^2}$

The function $G(x)$ is the integrand from the second integral:

$G(x) = \frac{1}{x^3}$

So, we have:

• $F(x) = -\frac{\ln(9x)}{2x^2}$
• $G(x) = \frac{1}{x^3}$

Would you like any further clarifications or details? Here are five follow-up questions to deepen your understanding:

1. How is the formula for integration by parts derived?
2. Why was $u = \ln(9x)$ chosen in this problem?
3. How do you determine which term to set as $u$ in integration by parts?
4. What would happen if we chose $u$ and $dv$ differently in this problem?
5. How would you solve this integral using substitution or another method?

Tip: When using integration by parts, always choose $u$ as the function that simplifies upon differentiation.