Anti derivatives of functions with Laurent series expansion

Question

I am given the functions $\mathrm{f}(z)=\frac{1}{z(1-z^2)}$ and $\mathrm{g}(z)=\frac{1}{z^2(1-z^2)}$. I am then asked to find the laurent series on the punctured disc with radius 1 and 0 as the center. I found this to be:
$f(z)=z^{-1}+\sum_{n=0}^{\infty}z^{2n+1}$ and $g(z)=z^{-2}+\sum_{n=0}^{\infty}z^{2n}$.
Then I am asked to argue if the functions have an anti derivative on the disc and if they have it on $\mathbb{C}\backslash\{-1,0,1\}$.
My first thought is that g has an anti derivative on the disc because we can find the anti derivative of all the terms. While f doesn't have one because we can't find one for the $z^{-1}$-term. but what about on $\mathbb{C}\backslash\{-1,0,1\}$? Do I have to find the laurent series outside the disc aswell, or is there a more direct way to see this? I feel like I am missing something.