indefinite integral problem

[Ashley5]

x^-3(x^2+4)dx I got u=x^2+4
du=2xdx
but how do you sub in for x^-3?
is it -3ln(u)? I don't get it

 

[soroban]

Hello, Ashley!

\(\displaystyle \int x^{-3}(x^2+4)\,dx\)

We have: .\(\displaystyle \displaystyle{\int x^{-3}\left(x^2 + 4\right)\,dx \;=\;\int\left(\frac{1}{x} + 4x^{-3}\right)\,dx} \;=\;\ln x - 2x^{-2} + C\)