To solve the integral \( \int 3x^2 \ln(2x^3) \, dx \), we can use integration by parts. Integration by parts is given by the formula:

\[ \int u \, dv = uv - \int v \, du \]

Here, we need to choose \( u \) and \( dv \) wisely. Let’s set:
\[ u = \ln(2x^3) \]
\[ dv = 3x^2 \, dx \]

Next, we need to find \( du \) and \( v \). To find \( du \), we differentiate \( u \) with respect to \( x \):

\[ u = \ln(2x^3) \]
\[ du = \frac{d}{dx} \ln(2x^3) \]
Using the chain rule:
\[ \ln(2x^3) = \ln(2) + \ln(x^3) = \ln(2) + 3\ln(x) \]
\[ \frac{d}{dx} \ln(2x^3) = \frac{d}{dx} (\ln(2) + 3\ln(x)) \]
\[ du = 3 \cdot \frac{1}{x} \, dx = \frac{3}{x} \, dx \]

Now, to find \( v \), we integrate \( dv \):
\[ dv = 3x^2 \, dx \]
\[ v = \int 3x^2 \, dx = x^3 \]

Now we have:
\[ u = \ln(2x^3) \]
\[ du = \frac{3}{x} \, dx \]
\[ dv = 3x^2 \, dx \]
\[ v = x^3 \]

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we get:
\[ \int 3x^2 \ln(2x^3) \, dx = x^3 \ln(2x^3) - \int x^3 \cdot \frac{3}{x} \, dx \]
\[ = x^3 \ln(2x^3) - \int 3x^2 \, dx \]
\[ = x^3 \ln(2x^3) - x^3 + C \]

where \( C \) is the constant of integration.

So, the final answer is:
\[ \int 3x^2 \ln(2x^3) \, dx = x^3 \ln(2x^3) - x^3 + C \]

Would you like more details or have any questions? 

Here are five more questions you might want to ask:
1. How do you integrate \( \int e^{x^2} \, dx \)?
2. Can you explain how to solve a differential equation?
3. What is the method of partial fractions in integration?
4. How do you find the Taylor series of a function?
5. What are the properties of definite integrals?

**Tip:** When using integration by parts, choose \( u \) and \( dv \) such that \( du \) simplifies the integral, making it easier to solve.

Learn how to integrate the function 3x^2 ln(2x^3) using integration by parts with a detailed step-by-step solution. This problem involves advanced calculus concepts including integration techniques and logarithmic functions.

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