I'm trying to show that $$\int_{|z|=r} \frac{\log z}{1+z^2} \ dz $$ goes to 0, as im taking $r \to 0 $
by the ML inequality $$\left| \int_{|z|=r} \frac{\log z}{1+z^2} \ dz \right| \leq \pi r \max |f(z)| $$ where $f(z)$ is the integrand in the first integral. My question is, since I'm taking $r \to 0$, do i have to calculate what $\max |f(z)| $ is to show that the integral is in fact 0?
You do not need to calculate it exactly. It is enough to have an upper bound. If $|z|=r>0$ then $$ \Bigl|\frac{\log z}{1+z^2}\Bigr|\le\frac{|\log r|+\pi}{1-r^2}. $$