If $C=\{z:|z|=r\} (r\neq 1)$, then show that $$\int_C \frac{dz}{1+z^2}= \begin{cases} -2\arctan r, & \text{if 0<r<1} \\[2ex] \pi - 2\arctan r, & \text{if 1<r} \end{cases}$$
This is an exercise from the chapter on Cauchy's Integral Formula. I do not know however, how to apply the formula to solve this. I know $\frac{1}{1+z^2}$ is a derivative of the $\arctan z$ function, but this has a branch cut inside $C$. I would greatly appreciate any help.