Why does $f:\mathbb{C}\setminus\{\pm 1\} \rightarrow\mathbb{C}; f(z)=\frac{1}{z^2-1}$ not have an antiderivative? My solution would be to define a counter-clockwise circle with the radius 1 around the point z=1 and integrate over it using the Residue-Theorem. So i get:
$\int\limits_{\partial B_1(1)}\frac{1}{z^2-1}\text{d}z=\int\limits_{\partial B_1(1)}\frac{1}{(z-1)(z+1)}\text{d}z=2\pi i\sum\limits_{a\in B_1(1)}$ind$_{\partial B_1(0)}(a)$ Res$_a(f)=2\pi i \frac{1}{2}=\pi i\neq 0$
As shown above, the value of the complex integral is not zero, and hence the function does not have an anti-derivative on this domain. Is this solution correct?

If you replace $B_1(0)$ by $B_1(1)$ everywhere, I agree.
To complete your argument: Suppose that there exists a holomorphic function $F:\mathbb C\setminus\{-1,1\}\to\mathbb C$ satisfying $F'=f$ on its domain. Let $\gamma:[0,1]\to\mathbb C$ be a parametrization of $\partial B_1(1)$ (for instance $\gamma(t)= 1+\exp(2\pi i t)$.)
Then
$$\pi i = \int_\gamma f =\int_0^1\frac{\mathrm d(F\circ\gamma)}{\mathrm dt}=F(\gamma(1))-F(\gamma(0))=0,$$
which is a contradiction.
