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.