Let $|\alpha|<1$ and $\psi_{\alpha}(z)=(\alpha-z)/(1-\bar\alpha z)$. I want to prove that $$\frac 1 {\pi} \int\int_{\mathbb{D}}|{\psi_{\alpha}}^{'}|dxdy = \frac{1-|\alpha|^2}{|\alpha|^2}\log\frac{1}{1-|\alpha|^2}$$
I calculated ${\psi_\alpha}^{'}(z)=(|\alpha|^2-1)/(1-\bar\alpha z)^2$. I substituted it and use $z=re^{i \theta}$ and need to integrate $1/|1-\bar\alpha re^{i \theta}|^2$ along circle of radius r (fixed). But how can I do this?
Convert to polars, as mentioned by Daniel Fischer. The integral over the disk becomes
$$\frac{1-|\alpha|^2}{\pi} \int_0^1 dr \, r \, \int_0^{2 \pi} \frac{d\theta}{|1-\bar{\alpha} r e^{i \theta}|^2} $$
Now,
$$|1-\bar{\alpha} r e^{i \theta}|^2 = (1-\bar{\alpha} r e^{i \theta})(1-\alpha r e^{-i \theta}) = 1+|\alpha|^2 r^2 - 2 \alpha_r r \cos{\theta} - 2 \alpha_i r \sin{\theta}$$
where $\alpha_r = \Re{(\alpha)}$ and $\alpha_i=\Im{(\alpha)}$. The integral over $\theta$ may be done via contour integration and the residue theorem. Let $z=e^{i \theta}$, then $d\theta=-i dz/z$, $\cos{\theta}=\frac12 (z+z^{-1})$ and $\sin{\theta}=(-i/2) (z-z^{-1})$. The integral over $\theta$ is then
$$i \oint_{|z|=1} \frac{dz}{\alpha r z^2 - (1+|\alpha|^2 r^2) z + \bar{\alpha} r}$$
Because we assume that $|\alpha| \lt 1$, the only pole of the above integrand for which we need compute the residue (i.e., inside the unit disk) is at $z=\bar{\alpha} r$. The value of the integral above is, by the residue theorem,
$$i 2 \pi \frac{-i}{2 r^2 |\alpha|^2 - 1 - r^2 |\alpha|^2} = \frac{2 \pi}{1-|\alpha|^2 r^2}$$
The integral over the unit disk is then
$$\frac{1-|\alpha|^2}{\pi} 2 \pi \int_0^1 dr \frac{r}{1-|\alpha|^2 r^2}= (1-|\alpha|^2) \int_0^1 \frac{du}{1-|\alpha|^2 u} $$
The sought-after result follows.
Without loss of generality, you can assume $0 < \alpha < 1$. Let us call $\rho := \alpha r$. Then
$$\frac{1}{\lvert 1 - \rho e^{i\theta}\rvert^2} = \frac{1}{1+\rho^2 - 2\rho\cos\theta},$$
and you can integrate that using one of the classical methods, like the residue theorem.
