The problem asks to use integration by parts to solve the integral:

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

We can solve this using the method of integration by parts, which is given by:

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

Step-by-step Solution:

1. Choose $u$ and $dv$:

   We set: $u = (\ln x)^2 \quad \text{so that} \quad du = 2 \ln x \cdot \frac{1}{x} \, dx$ And, $dv = dx \quad \text{so that} \quad v = x$

2. Apply the Integration by Parts Formula:

   Using the formula $\int u \, dv = uv - \int v \, du$, we have: $\int (\ln x)^2 \, dx = x (\ln x)^2 - \int x \cdot 2 \ln x \cdot \frac{1}{x} \, dx$ Simplify the second term: $\int x \cdot 2 \ln x \cdot \frac{1}{x} \, dx = \int 2 \ln x \, dx$

3. Solve $\int 2 \ln x \, dx$:

   Again, we apply integration by parts for $\int 2 \ln x \, dx$.

   Let: $u = \ln x \quad \text{so that} \quad du = \frac{1}{x} \, dx$ And, $dv = dx \quad \text{so that} \quad v = x$

   Now apply the formula: $\int 2 \ln x \, dx = 2 (x \ln x - \int x \cdot \frac{1}{x} \, dx)$ Simplify: $\int 2 \ln x \, dx = 2 (x \ln x - x) = 2x \ln x - 2x$

4. Substitute Back:

   Now substitute back into the original expression: $\int (\ln x)^2 \, dx = x (\ln x)^2 - (2x \ln x - 2x) + C$ Simplify: $\int (\ln x)^2 \, dx = x (\ln x)^2 - 2x \ln x + 2x + C$

Thus, the integral is:

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

Would you like further clarification or additional details on any steps?

Here are some related questions to explore:

1. How does integration by parts generally work?
2. Why was $u = (\ln x)^2$ chosen in this case?
3. What other methods can be used to integrate functions involving $\ln x$?
4. How can integration by parts be extended to higher powers of $\ln x$?
5. What are the applications of such logarithmic integrals in physics or engineering?

Tip: When applying integration by parts, always choose $u$ and $dv$ carefully. Selecting $u$ as the part that simplifies upon differentiation often helps streamline the solution.