I am trying to compute the following integral $$\int_0^1 r dr \int_0^{2\pi} d\theta \frac{1+r e^{i \theta}}{\left(1-r e^{i \theta}\right)^3}f(r,\theta) $$ assuming that the function $f$ is regular in the disk. I expect this integral to be related to $f(1,0)$ but I have no idea of how to prove it.
Thank you for all your help.
Taking $f(r,\theta)$ to mean $f(re^{i\theta})$, the inner integral is the same as the contour integral
$$\oint_{|z| = r} \frac{dz}{iz} \frac{1 + z}{(1 - z)^3}f(z)$$
Since $0 < r < 1$, the function $\frac{1+z}{i(1 - z)^3}f(z)$ is regular inside and on the circle $|z| = r$. By the Cauchy integral formula this integral evaluates to $$2\pi i\frac{1+0}{i(1-0)^3}f(0) = 2\pi f(0)$$ Your integral reduces to $\int_0^1 2\pi f(0)\, r\, dr = \pi f(0)$.