Cauchy Theorem: $f$ is analytic for $|z|<2$ and $\alpha \in \mathbb{C}$ evaluate $I=\int_{|z|=1} (Re z+\alpha)\frac{f(z)}{z}dz.$

Question

Suppose that $f$ is analytic for $|z|<2$ and $\alpha$ is a complex constant. Evaluate $$I=\int_{|z|=1} (Re z+\alpha)\frac{f(z)}{z}dz.$$

On $|z|=1$, $Re z=\frac{1}{2}(z+\frac{1}{z})$ so we can get $\frac{1}{2}\int_{|z|=1}(z^2+1+2\alpha z)\frac{f(z)}{z^2}dz$ in an attempt to get rid of $Re z$, but I do not know how to proceed from here. This is from a chapter on Cauchy's Theorem and I would greatly appreciate any help.