To solve the integral $\int \left(\ln(x)\right)^2 \, dx$, we can use integration by parts. Let's go through the solution step by step.

Step 1: Set up integration by parts

We know the formula for integration by parts: $\int u \, dv = uv - \int v \, du$ Here, let's choose: $u = (\ln(x))^2 \quad \text{and} \quad dv = dx$

Step 2: Differentiate $u$ and integrate $dv$

First, differentiate $u$: $du = 2\ln(x) \cdot \frac{1}{x} \, dx = \frac{2\ln(x)}{x} \, dx$ Next, integrate $dv$: $v = \int dx = x$

Step 3: Apply the integration by parts formula

Now, substitute into the integration by parts formula: $\int (\ln(x))^2 \, dx = x(\ln(x))^2 - \int x \cdot \frac{2\ln(x)}{x} \, dx$ This simplifies to: $\int (\ln(x))^2 \, dx = x(\ln(x))^2 - 2\int \ln(x) \, dx$

Step 4: Solve $\int \ln(x) \, dx$

We now need to solve the integral $\int \ln(x) \, dx$. We can again use integration by parts, with: $u = \ln(x) \quad \text{and} \quad dv = dx$ Differentiate and integrate as before: $du = \frac{1}{x} \, dx \quad \text{and} \quad v = x$ Apply the integration by parts formula: $\int \ln(x) \, dx = x\ln(x) - \int x \cdot \frac{1}{x} \, dx = x\ln(x) - \int 1 \, dx = x\ln(x) - x$

Step 5: Substitute back into the original equation

Substitute the result of $\int \ln(x) \, dx$ back into the original equation: $\int (\ln(x))^2 \, dx = x(\ln(x))^2 - 2\left(x\ln(x) - x\right) + C$ Simplify the expression: $\int (\ln(x))^2 \, dx = x(\ln(x))^2 - 2x\ln(x) + 2x + C$

Thus, the final answer is: $\boxed{\int (\ln(x))^2 \, dx = x(\ln(x))^2 - 2x\ln(x) + 2x + C}$

Would you like any further details or explanations? Here are some additional related questions to explore:

1. How would you apply integration by parts to $\int x \ln(x) \, dx$?
2. What is the integral of $\int \ln(x) \, dx$ if you change the variable $x$ to $e^t$?
3. How does the integral change if $(\ln(x))^2$ is replaced with $(\ln(x))^3$?
4. Can you apply a substitution method to solve $\int \left(\ln(x)\right)^2 \, dx$?
5. What are the general steps for choosing $u$ and $dv$ in integration by parts?

Tip: Always double-check the constants and coefficients when using integration by parts, as small errors can propagate through the calculation.