Does $f(z) = \dfrac{1}{z-1} - \dfrac{1}{z+1}$ have an antiderivative in $|z|>2?$
I know the condition for $f$ having an antiderivative is $\int_{\gamma} f(z)\,dz = 0$.
I'm not sure whether it has an antiderivative or not. I tried taking some circle $C$ in $|z|>2$ and showing $\int_{C} f(z)\,dz \neq 0$ but didn't succeed. If it does have an antiderivative, I'm not sure how to go about proving $\int_{\gamma} f(z)\,dz = 0$ for any $\gamma$.
If $\gamma$ is a path in $\{z\in\mathbb C\mid\lvert z\rvert>2\}$, then $\operatorname{ind}(\gamma;1)=\operatorname{ind}(\gamma;-1)$. Therefore, by the residue thorem,$$\int_\gamma\frac{\mathrm dz}{z-1}-\int_\gamma\frac{\mathrm dz}{z+1}=2\pi i\bigl(\operatorname{ind}(\gamma;1)-\operatorname{ind}(\gamma;-1)\bigr)=0.$$