Different answers on $\frac{1}{2\pi}\int_0^{2\pi} \frac{1}{e^{ix}+c}\ dx$ where $c\in\mathbb{C}$ is fixed Upon the substitution $z=e^{ix}$, I get that this is equal to
$$\frac{1}{2\pi i} \oint_{C} \frac{dz}{z(z+c)},$$
where the integral is over the unit circle. I then split this up into
$$\frac{1}{2c\pi i}\left(\oint_{C} \frac{dz}{z}-\oint_{C}\frac{1}{z+c}\right)dz.$$
From here, I get that the first integral is $2\pi i$ and the second integral is $2\pi i$ if $c$ is inside the unit circle and $0$ if $c$ is outside. From this question, it seems as if the second integral should be $\pi i$ if $c$ is on the unit circle itself. Thus, I get that this integral should be
$$f(c)=\frac{1}{2\pi}\int_0^{2\pi} \frac{1}{e^{ix}+c}\ dx = \begin{cases} 0 & \mathrm{if}\ |c|<1 \\ \frac{1}{2c} & \mathrm{if} \ |c|=1 \\ \frac{1}{c}  & \mathrm{otherwise.}\end{cases}$$
However, upon experimentation with various values of $c$ on the Online Integral Calculator, it seems as if
$$f(c)=\begin{cases}0 & \mathrm{if}\ c=0 \\ \frac{3}{2} & \mathrm{if} \ c=1 \\ \frac{1}{c} & \mathrm{otherwise.}\end{cases}$$
What's going on?

If $c=0$, then
$$\frac{1}{2\pi}\int_{0}^{2\pi}\frac{dx}{e^{ix}+c}dx=\frac{1}{2\pi}\int_{0}^{2\pi}e^{-ix}dx = 0.
$$
If $c\ne 0$, then
\begin{align}
   \frac{1}{2\pi}\int_{0}^{2\pi}\frac{dx}{e^{ix}+c}&=\frac{1}{2\pi}\int_{0}^{2\pi}\frac{\frac{1}{ie^{ix}}d(e^{ix})}{e^{ix}+c} \\
  &= \frac{1}{2\pi i}\int_{|z|=1} \frac{dz}{z(z+c)} \\
  &=\frac{1}{2\pi ic}\int_{|z|=1}\frac{1}{z}-\frac{1}{z+c}dz.
\end{align}
If $|c| < 1$ the above is $0$. If $c > 1$, the above is $\frac{1}{c}$.
