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The given problem asks for an expression that is equivalent to the integral

\[
\int \left(\ln x\right)^{1/2} \ln (\ln x) \, dx
\]

after performing an appropriate substitution.

### Step-by-step process:

1. **Substitution**: Start by substituting \( u = \ln x \). Then, the derivative of \( u \) is:

   \[
   du = \frac{1}{x} dx
   \]
   This implies that \( dx = x \, du = e^u \, du \).

2. **Rewriting the Integral**: After the substitution, the expression becomes:

   \[
   \int \left(u^{1/2} \ln u\right) e^u \, du
   \]

3. **Simplifying the Expression**: Now we have an integral in terms of \( u \), and the rest of the steps involve integration by parts or further simplification to match one of the options.

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Let’s analyze the given answer choices based on this substitution framework:

- **First option**: \( \frac{u^2(\ln u)^2}{2} - \frac{u^2 \ln u}{2} + \frac{1}{2} \int u \, du + C \)

- **Second option**: \( 2u^{2/3} \ln u - \frac{2}{3} \int u \, du + C \)

- **Third option**: \( 2u^{3/2} \ln u - \frac{2}{3} \int u \, du + C \)

The correct expression after solving should be:

\[
2u^{3/2} \ln u - \frac{2}{3} \int u \, du + C
\]

This matches the **third option**.

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Do you want more details on any specific step?

### Related Questions:
1. What is the integral of \( u^{1/2} \ln u \)?
2. How does integration by parts apply to this problem?
3. What are common substitutions used in logarithmic integrals?
4. How do you handle complex logarithmic integrals in general?
5. Why is substitution a helpful technique in integrals involving logarithms?

**Tip**: When integrating functions involving logarithms, substitutions like \( u = \ln x \) simplify the expressions significantly and often lead to integrable forms.

Which of the following expression is equivalent to the integral \( \int (\ln x)^{1/2} \ln (\ln x) \, dx \) after appropriate substitution?

Learn how to solve the integral \( \int (\ln x)^{1/2} \ln (\ln x) \, dx \) using substitution and integration techniques. This problem involves logarithmic functions and substitution, suitable for advanced calculus students.

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