The problem in the image asks to evaluate the integral

$\int \frac{\ln(9x)}{x^6} \, dx$

and provides a clue: it suggests using integration by parts, with $u = \ln(9x)$.

The integration by parts formula is:

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

Step 1: Define $u$ and $dv$

Let:

• $u = \ln(9x)$, which implies $du = \frac{1}{x} \, dx$,
• $dv = \frac{1}{x^6} \, dx$, which implies $v = -\frac{1}{5x^5}$.

Step 2: Apply integration by parts

Using the formula $\int u \, dv = uv - \int v \, du$, we substitute:

$\int \frac{\ln(9x)}{x^6} \, dx = \ln(9x) \cdot \left(-\frac{1}{5x^5}\right) - \int \left(-\frac{1}{5x^5}\right) \cdot \frac{1}{x} \, dx$

This will simplify into the sum of two functions: $F(x)$ and the integral of $G(x) \, dx$.

Step 3: Identify $F(x)$ and $G(x)$

From the integration by parts result:

• $F(x)$ is the term $\ln(9x) \cdot \left(-\frac{1}{5x^5}\right) = -\frac{\ln(9x)}{5x^5}$,
• $G(x)$ is the function in the remaining integral, which is $\frac{1}{5x^6}$.

Thus:

• $F(x) = -\frac{\ln(9x)}{5x^5}$,
• $G(x) = -\frac{1}{5x^6}$.

Would you like more details on the integration process or have any further questions?

Here are 5 related questions for further exploration:

1. How is the integration by parts rule derived?
2. What are some practical applications of integration by parts?
3. How would this integral change if $\ln(9x)$ were replaced with $\ln(x)$?
4. What happens if we reverse the roles of $u$ and $dv$ in integration by parts?
5. Can the integration by parts method be applied to definite integrals?

Tip: When performing integration by parts, always choose $u$ such that its derivative simplifies the integral.