To integrate $(\ln(y))^2$ with respect to $y$, we use the method of integration by parts.

The integral is:

$I = \int (\ln(y))^2 \, dy$

Step 1: Set up integration by parts

Let:

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

The formula for integration by parts is: $\int u \, dv = uv - \int v \, du$

Step 2: Apply the formula

$\int (\ln(y))^2 \, dy = y (\ln(y))^2 - \int y \cdot 2 \ln(y) \cdot \frac{1}{y} \, dy$ Simplify the second term: $\int (\ln(y))^2 \, dy = y (\ln(y))^2 - 2 \int \ln(y) \, dy$

Step 3: Solve $\int \ln(y) \, dy$

The integral of $\ln(y)$ is well-known: $\int \ln(y) \, dy = y (\ln(y) - 1)$

Step 4: Substitute back

Substitute $\int \ln(y) \, dy = y (\ln(y) - 1)$ into the equation: $\int (\ln(y))^2 \, dy = y (\ln(y))^2 - 2 \big[y (\ln(y) - 1)\big]$ Expand and simplify: $\int (\ln(y))^2 \, dy = y (\ln(y))^2 - 2y \ln(y) + 2y$

Final Answer:

$\int (\ln(y))^2 \, dy = y \big((\ln(y))^2 - 2 \ln(y) + 2\big) + C$

Let me know if you'd like further clarification or have more questions!

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Related Questions:

1. How do you apply integration by parts to more complex functions?
2. What is the general formula for integrating powers of logarithms?
3. Can $\int (\ln(x))^n dx$ be solved for any $n$?
4. How does the integral of $\ln(x)$ differ when involving definite limits?
5. What are practical applications of integrating logarithmic powers?

Tip:

For powers of logarithmic terms, consider patterns in integration by parts to simplify repetitive steps.