Show that $f(z)$ has no antiderivative in $\,S=\mathbb{C}\setminus \{-i,i\}$ $f(z)=\frac{1}{z^{2}+1}$
I know that you can do this using a proof by contradiction and by showing that if you assume it has an anti-derivative that it wouldn't follow the fundamental theorem of calculus which would be a contradiction but I don't know how to show this.
 A: Edited since later comments made it clear that the OP wanted another solution.
First solution (some theory, very little computation)
Once you know the residue theorem and some methods to compute residues, the slick argument is this: $\operatorname{Res}(f;\pm i) = \pm\frac{1}{2i}$ which means that the integral of $f$ along a small closed curve around either pole is non-zero.
Second solution (a litte less theory, a little more computation)
If you don't know anything about residues, but you still know Cauchy's integral formula, you can argue like this:
Let $\gamma$ be a small curve going once around $z=i$ (and not around $z=-i$). Then
$$\int_\gamma \frac{dz}{z^2+1} = \int_\gamma \dfrac{\frac{1}{z+i}}{z-i}\,dz = 2\pi i \cdot \frac{1}{2i} \neq 0,$$
with a similar computation for a small curve around $z=-i$.
Third solution (very little theory, lots of computation)
If you don't know Cauchy's integral formula yet, you need to work harder.
Let $\gamma$ be a small circle going once around $z=i$ (and not around $z=-i$). Do a partial fractions decomposition to split your integral into two pieces. Parametrize the circle, and compute both integrals (one of them preferably via Cauchy's integral theorem if you know it). Repeat for a small circle around $z=-i$.
A: Use that $z^2+1=(z+i) (z-i)$ and partial fraction decomposition.
In fact 
$$\frac{1}{z^2+1} = \frac{i}{2(i+z)}-\frac{i}{2(z-i)}$$
So $\frac{1}{z^2+1}$ has an antriderivative on $\mathbb{C}\setminus\{i,-i\}$ iff
$\frac{1}{z}$ has an antiderivative on $\mathbb{C}\setminus\{0\}$ 
