https://file.aiquickdraw.com/mathbot/file/1726714160894zsv7r97y.jpg

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.

Consider the integral: ∫(ln(9x)/x^6) dx. The problem involves using the integration by parts technique, assigning u = ln(9x).

Learn how to evaluate the integral ∫(ln(9x)/x^6) dx by applying the integration by parts technique. This solution covers logarithmic differentiation and key calculus concepts. Ideal for university-level calculus students.

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