To solve the integral $I = \int x^2 \ln(2x) \, dx$ using integration by parts, we will apply the standard formula for integration by parts:

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

We need to choose parts of the integral for $u$ and $dv$. Typically, we set $u$ to the logarithmic part and $dv$ to the remaining part, as differentiating the logarithmic term simplifies the problem.

Step 1: Assign $u$ and $dv$

Let:

• $u = \ln(2x)$, so $du = \frac{1}{x} \, dx$
• $dv = x^2 \, dx$, so $v = \frac{x^3}{3}$

Step 2: Apply the integration by parts formula

Using the formula $\int u \, dv = u v - \int v \, du$, we substitute the values:

$I = \left( \ln(2x) \cdot \frac{x^3}{3} \right) - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx$

Simplify the integral:

$I = \frac{x^3}{3} \ln(2x) - \frac{1}{3} \int x^2 \, dx$

Step 3: Solve the remaining integral

Now, solve $\int x^2 \, dx$:

$\int x^2 \, dx = \frac{x^3}{3}$

Substitute this back into the equation:

$I = \frac{x^3}{3} \ln(2x) - \frac{1}{3} \cdot \frac{x^3}{3}$

$I = \frac{x^3}{3} \ln(2x) - \frac{x^3}{9}$

Step 4: Write the final answer

Thus, the integral is:

$I = \frac{x^3}{3} \ln(2x) - \frac{x^3}{9} + C$

where $C$ is the constant of integration.

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Would you like further clarification on any of the steps?

Here are five related questions you can explore:

1. How does integration by parts work for other logarithmic functions?
2. What if the integral was $\int x^3 \ln(2x) \, dx$?
3. How can you derive the integration by parts formula?
4. What happens if you change the base of the logarithm in the integral?
5. Can integration by parts be applied to products of trigonometric and logarithmic functions?

Tip: Always choose $u$ as the function that simplifies upon differentiation when using integration by parts.