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Do questions 23, 25 and show all the work please

Let's solve each of these integrals step-by-step.

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### Problem 23

Evaluate the improper integral:
\[
\int_2^{\infty} \frac{1}{x \ln x} \, dx
\]

#### Solution:

1. **Set Up the Integral with a Limit**:
   Since the integral has an upper bound of infinity, we rewrite it as a limit:
   \[
   \int_2^{\infty} \frac{1}{x \ln x} \, dx = \lim_{b \to \infty} \int_2^b \frac{1}{x \ln x} \, dx
   \]

2. **Substitute**:
   Let \( u = \ln x \). Then \( du = \frac{1}{x} dx \).
   When \( x = 2 \), \( u = \ln 2 \).
   When \( x = b \), \( u = \ln b \).

   Substituting into the integral, we get:
   \[
   \int_2^b \frac{1}{x \ln x} \, dx = \int_{\ln 2}^{\ln b} \frac{1}{u} \, du
   \]

3. **Integrate**:
   The integral of \( \frac{1}{u} \) is \( \ln |u| \):
   \[
   \int_{\ln 2}^{\ln b} \frac{1}{u} \, du = \left[ \ln |u| \right]_{\ln 2}^{\ln b} = \ln(\ln b) - \ln(\ln 2)
   \]

4. **Take the Limit**:
   Now we take the limit as \( b \to \infty \):
   \[
   \lim_{b \to \infty} \left( \ln(\ln b) - \ln(\ln 2) \right)
   \]
   As \( b \to \infty \), \( \ln b \to \infty \), so \( \ln(\ln b) \to \infty \) as well. This means that the integral diverges.

#### Conclusion:
\[
\int_2^{\infty} \frac{1}{x \ln x} \, dx \text{ diverges.}
\]

---

### Problem 25

Evaluate the improper integral:
\[
\int_0^{\infty} x e^{4x} \, dx
\]

#### Solution:

1. **Set Up the Integral with a Limit**:
   Similar to before, we express this as a limit:
   \[
   \int_0^{\infty} x e^{4x} \, dx = \lim_{a \to \infty} \int_0^a x e^{4x} \, dx
   \]

2. **Integration by Parts**:
   To integrate \( x e^{4x} \), use integration by parts:
   - Let \( u = x \) and \( dv = e^{4x} dx \).
   - Then \( du = dx \) and \( v = \frac{e^{4x}}{4} \).

   Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
   \[
   \int x e^{4x} \, dx = \frac{x e^{4x}}{4} - \int \frac{e^{4x}}{4} \, dx
   \]

3. **Integrate the Remaining Integral**:
   Now, integrate \( \int \frac{e^{4x}}{4} \, dx \):
   \[
   \int \frac{e^{4x}}{4} \, dx = \frac{e^{4x}}{16}
   \]

   Putting it all together:
   \[
   \int x e^{4x} \, dx = \frac{x e^{4x}}{4} - \frac{e^{4x}}{16}
   \]

4. **Evaluate the Definite Integral**:
   Substitute the limits \( 0 \) and \( a \):
   \[
   \int_0^a x e^{4x} \, dx = \left( \frac{a e^{4a}}{4} - \frac{e^{4a}}{16} \right) - \left( \frac{0 \cdot e^{0}}{4} - \frac{e^{0}}{16} \right)
   \]
   Simplify:
   \[
   = \frac{a e^{4a}}{4} - \frac{e^{4a}}{16} + \frac{1}{16}
   \]

5. **Take the Limit**:
   Now, take the limit as \( a \to \infty \):
   \[
   \lim_{a \to \infty} \left( \frac{a e^{4a}}{4} - \frac{e^{4a}}{16} + \frac{1}{16} \right)
   \]
   As \( a \to \infty \), \( e^{4a} \) grows faster than any polynomial term, so \( \frac{a e^{4a}}{4} \) and \( \frac{e^{4a}}{16} \) both approach infinity. This means that the integral diverges.

#### Conclusion:
\[
\int_0^{\infty} x e^{4x} \, dx \text{ diverges.}
\]

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Let me know if you need any clarifications or further details!

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#### Related Questions

1. How does substitution work in improper integrals?
2. What is the role of limits in evaluating improper integrals?
3. How do exponential terms affect convergence or divergence of an integral?
4. When should we use integration by parts?
5. Can integration by parts be used iteratively for complex expressions?

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#### Tip

For integrals with terms that grow exponentially, always check if they converge by analyzing the behavior of the integrand as \( x \to \infty \).

Explore detailed solutions for improper integrals, including substitution and integration by parts, in questions 23 and 25. Learn to assess divergence in integrals such as \( \int_2^{\infty} \frac{1}{x \ln x} dx \) and \( \int_0^{\infty} x e^{4x} dx \). Suitable for college calculus students.

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