Is there an anti derivative to $f(z)=\frac{z}{(z-1)(z-2)}$ in $|z|>4$?

Question

Is there an anti derivative function to the complex function $$f(z)=\frac{z}{(z-1)(z-2)}$$ in $|z|>4$?

My attempt: We notice that $f(z)=\frac{z}{z-2}+\frac{z}{1-z}$ (by decomposing to partial fractions). But when I try to guess an anti-derivative function it doesn't go well: $$ (\ln(\frac{1}{1-z})'=\frac{1}{1-x} \\(\ln\frac{z}{1-z})'=\frac{1}{x(1-x)} $$

Answer 1

No, because$$\int_{\lvert z\rvert=5}f(z)\,\mathrm dz=2\pi i,$$by the residue theorem. If there was such an antiderivative, the integral along any loop would be equal to $0$.