It looks like the image contains a mathematical integral to solve:

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

Let's proceed with solving this integral using integration by parts.

Step 1: Use Integration by Parts

We start by setting up the formula for integration by parts:

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

Let:

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

Step 2: Apply the Formula

Now, applying the integration by parts:

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

Simplifying the second integral:

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

Step 3: Solve $\int \ln(x) \, dx$

We now use integration by parts again for the integral $\int \ln(x) \, dx$.

• Let $u = \ln(x)$, so $du = \frac{1}{x} \, dx$
• Let $dv = dx$, so $v = x$

This gives:

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

Step 4: Substitute Back

Now, substitute this result into our original expression:

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

Simplifying further:

$\int \ln^2(x) \, dx = x \ln^2(x) - 2x \ln(x) + 2x + C$

Final Answer:

The result of the integral is:

$x \ln^2(x) - 2x \ln(x) + 2x + C$

Now, let's match it with the options given in your image. It looks like the correct choice is:

Option (a): $x \ln^2(x) - 2x \ln(x) + 2x + C$

Would you like more details on the steps or have any further questions about the solution? Here are some related questions you might be interested in:

1. How do you use integration by parts for other types of logarithmic integrals?
2. Can you explain the general formula for integration by parts in more detail?
3. How can we solve integrals involving higher powers of logarithms?
4. What would happen if the integral involved $\ln(x)^n$ for $n > 2$?
5. How do we approach integrals that have both polynomial and logarithmic terms?

Tip: When solving logarithmic integrals, breaking them down into parts and using substitution or integration by parts repeatedly often helps simplify the process.