To compute the integral of $\ln(y) \, dy$, we use the method of integration by parts.

Formula for Integration by Parts:

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

Here, we let:

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

Applying the formula:

$\int \ln(y) \, dy = y \ln(y) - \int y \cdot \frac{1}{y} \, dy$

Simplify the second term: $\int \ln(y) \, dy = y \ln(y) - \int 1 \, dy$

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

Final Answer:

$\int \ln(y) \, dy = y (\ln(y) - 1) + C$

Where $C$ is the constant of integration.

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Related Questions:

1. What is the derivative of $y(\ln(y) - 1)$?
2. How do you apply integration by parts to other logarithmic integrals?
3. How can this integral be applied in real-world problems involving logarithmic growth?
4. What happens if the integrand is $(\ln(y))^2$?
5. Can this method extend to $\ln(f(y))$, where $f(y)$ is a differentiable function?

Tip: Integration by parts is particularly useful when a product of functions is involved, especially logarithmic or inverse trigonometric functions!