Question

How do I evaluate the integral `int7x^2 ln(x) dx`?

Answer 1

I'd use integration by parts: since `(fg)'=f'g+fg'`, by integrating this relation one has that `fg=\int f'g + \int fg'`.

When you have to integrate the product of two functions, integration by parts is often a good strategy: if one of the two factors is easily integrable/derivable, you can try and see if the integral becomes easier. In your case, I'd use the fact that
`(1) \int f'g=fg-\int fg'`

So, we have the following:
`f(x)= 7/3 x^3,\ \ \ f'(x)= 7x^2`
`g(x)=\ln(x),\ \ \ g'(x)=1/x`.

And using the equation labeled as `(1)`, you have that

`\int 7x^2 \ln(x) = 7/3 x^3 \ln(x) - \int 7/3 x^3 1/x`

The andantage is evident: the integral of `7/3 x^3 1/x=7/3 x^2` is obviously `7/9 x^3`, and so our solution is

`\int 7x^2 \ln(x) = 7/3 x^3 \ln(x) - 7/9 x^3+c`