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The problem asks for the values of \( p \) for which the integral

\[
\int_2^{\infty} \frac{1}{x (\ln x)^p} \, dx
\]

converges.

To determine this, we will analyze the integral at its potential points of divergence, specifically at \( x = \infty \). The behavior near \( x = 2 \) is not problematic, as the integrand does not blow up. Therefore, the main issue is with \( x \to \infty \).

### Step-by-Step Approach:

1. **Substitution to Simplify**:
   Let’s use the substitution:
   \[
   u = \ln x \quad \text{so that} \quad du = \frac{1}{x} dx.
   \]
   This changes the bounds of integration:
   \[
   \text{When } x = 2, \ u = \ln 2.
   \quad \text{When } x \to \infty, \ u \to \infty.
   \]
   The integral becomes:
   \[
   \int_{\ln 2}^{\infty} \frac{du}{u^p}.
   \]

2. **Analyzing Convergence**:
   The behavior of the integral depends on the value of \( p \).

   - For \( p = 1 \), the integral becomes \( \int_{\ln 2}^{\infty} \frac{du}{u} \), which is a divergent logarithmic integral.
   - For \( p > 1 \), the integral becomes \( \int_{\ln 2}^{\infty} u^{-p} \, du \), which converges because the integral of \( u^{-p} \) for \( p > 1 \) yields a finite result.
   - For \( p < 1 \), the integral behaves like \( u^{1-p} \), which diverges as \( u \to \infty \) since the exponent \( 1-p > 0 \).

### Conclusion:
The integral converges for \( p > 1 \).

Would you like further explanation or clarification on any step?

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Here are 5 related questions to deepen your understanding:

1. How does the substitution \( u = \ln x \) simplify the analysis of the integral?
2. What happens to the integral when \( p = 1 \), and why does it diverge?
3. Could you explore what happens if the lower bound of the integral changes, for example, to 1 instead of 2?
4. What is the general criterion for convergence for integrals of the form \( \int_1^\infty \frac{1}{x^a} dx \)?
5. How can you relate this problem to the comparison test for improper integrals?

**Tip**: Whenever solving improper integrals, always examine the behavior near points of infinity or where the function could become unbounded.

Find the values of p for which the integral ∫ from 2 to ∞ of 1/(x (ln x)^p) dx converges.

Determine the values of p for which the improper integral ∫ 1/(x (ln x)^p) dx from 2 to ∞ converges. This solution covers the substitution method and analyzes the integral's convergence behavior for different values of p. Suitable for undergraduate calculus students.

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