To solve the integral:

$I = \int x (\ln(x))^2 \, dx$

We'll use integration by parts. Let’s follow the standard method of breaking the integral into two parts: one part to differentiate, and the other to integrate.

Step 1: Set up for Integration by Parts

Recall the formula for integration by parts:

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

For our integral, let’s choose:

• $u = (\ln(x))^2$, so that $du = 2 \ln(x) \cdot \frac{1}{x} \, dx$.
• $dv = x \, dx$, so that $v = \frac{x^2}{2}$.

Step 2: Apply the formula

Now, applying the integration by parts formula:

$I = \int x (\ln(x))^2 \, dx = \frac{x^2}{2} (\ln(x))^2 - \int \frac{x^2}{2} \cdot 2 \ln(x) \cdot \frac{1}{x} \, dx$

Simplifying the second integral:

$I = \frac{x^2}{2} (\ln(x))^2 - 2 \int x \ln(x) \, dx$

Step 3: Solve the remaining integral

We still need to solve $\int x \ln(x) \, dx$. We can use integration by parts again for this integral. Let:

• $u = \ln(x)$, so that $du = \frac{1}{x} \, dx$.
• $dv = x \, dx$, so that $v = \frac{x^2}{2}$.

Now applying the integration by parts formula:

$\int x \ln(x) \, dx = \frac{x^2}{2} \ln(x) - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx$

Simplifying the remaining integral:

$\int x \ln(x) \, dx = \frac{x^2}{2} \ln(x) - \frac{x^2}{4}$

Step 4: Substitute back

Substitute this result back into the expression for $I$:

$I = \frac{x^2}{2} (\ln(x))^2 - 2 \left( \frac{x^2}{2} \ln(x) - \frac{x^2}{4} \right)$

Simplify the terms:

$I = \frac{x^2}{2} (\ln(x))^2 - \left( x^2 \ln(x) - \frac{x^2}{2} \right)$

Step 5: Final simplification

$I = \frac{x^2}{2} (\ln(x))^2 - x^2 \ln(x) + \frac{x^2}{2} + C$

Thus, the solution to the integral is:

$\boxed{ I = \frac{x^2}{2} (\ln(x))^2 - x^2 \ln(x) + \frac{x^2}{2} + C }$

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Do you have any questions or need further details?

Here are 5 related questions:

1. How is integration by parts applied in general?
2. What other techniques are useful for logarithmic integrals?
3. Can you explain how to differentiate logarithmic functions?
4. What are the common cases where integration by parts is preferred?
5. How does this technique extend to integrals involving powers of logarithms?

Tip: When applying integration by parts, always simplify as much as possible after each step to make the next part easier!