To evaluate the integral $\int 2x^5 \ln(5x^2) \, dx$ using integration by parts, we are given: $u = \ln(5x^2)$ $dv = 2x^5 \, dx$

We need to find $du$ and $v$.

First, let's find $du$: $u = \ln(5x^2)$ To differentiate $u$, we use the chain rule: $\ln(5x^2) = \ln(5) + \ln(x^2) = \ln(5) + 2\ln(x)$ $\frac{d}{dx} \ln(5x^2) = \frac{d}{dx} (\ln(5) + 2\ln(x))$ $du = 2 \cdot \frac{1}{x} \, dx = \frac{2}{x} \, dx$

Now, let's find $v$: $dv = 2x^5 \, dx$ Integrating $dv$ with respect to $x$: $v = \int 2x^5 \, dx = \frac{2}{6} x^6 = \frac{1}{3} x^6$

Now we have: $u = \ln(5x^2)$ $du = \frac{2}{x} \, dx$ $dv = 2x^5 \, dx$ $v = \frac{1}{3} x^6$

Using the integration by parts formula $\int u \, dv = uv - \int v \, du$, we get: $\int 2x^5 \ln(5x^2) \, dx = \ln(5x^2) \cdot \frac{1}{3} x^6 - \int \left( \frac{1}{3} x^6 \cdot \frac{2}{x} \right) dx$ $= \frac{1}{3} x^6 \ln(5x^2) - \int \frac{2}{3} x^5 \, dx$ $= \frac{1}{3} x^6 \ln(5x^2) - \frac{2}{3} \int x^5 \, dx$ $= \frac{1}{3} x^6 \ln(5x^2) - \frac{2}{3} \cdot \frac{1}{6} x^6$ $= \frac{1}{3} x^6 \ln(5x^2) - \frac{1}{9} x^6 + C$

Therefore, the final answer is: $\int 2x^5 \ln(5x^2) \, dx = \frac{1}{3} x^6 \ln(5x^2) - \frac{1}{9} x^6 + C$

Would you like more details or have any questions?

Here are five more questions you might want to ask:

1. How do you integrate $\int x e^x \, dx$ using integration by parts?
2. Can you explain how to solve a second-order differential equation?
3. What is the method of substitution in integration?
4. How do you find the Maclaurin series of a function?
5. What are the properties of improper integrals?

Tip: When dealing with logarithmic functions in integration by parts, make sure to carefully handle the argument of the logarithmic function to avoid mistakes in differentiation.