I'm trying to learn how to calculate $$ \oint_\gamma \frac{\ln(\overline{z} - b)}{z - a} dz$$
using the Cauchy-Pompeiu formula:
$$ f(a) = \frac{1}{2\pi i}\int_{\gamma} \frac{f(z) \,dz}{z-a} - \frac{1}{\pi}\iint_D \frac{\partial f}{\partial \bar{z}}(z) \frac{dx\wedge dy}{z-a}$$
where $D$ is the area enclosed by $\gamma$, $\gamma$ is a simple closed curve in the complex plane, $a \in D$, $b \notin D$ and I'm taking the principal branch of the logarithm. I figure I can set $f(z) = \ln(\overline{z} - b)$ and go from there.
To calculate $\frac{\partial f}{\partial \bar{z}}(z)$ I have
$$\begin{align} \frac{\ln(\overline{z} - b)}{\partial \bar{z}}(z) &= \frac{1}{2} (\frac{\partial \ln(x - iy - b)}{\partial x} + i \frac{\partial \ln(x - iy - b)}{\partial y}) \\ &= \frac{1}{2} (\frac{1}{x - iy - b} + i \frac{-i}{x - iy - b} ) \\ &= \frac{1}{\overline{z} - b} \end{align}$$
So then $$ \int_{\gamma} \frac{\ln(\overline{z} - b)}{z-a} dz = 2 \pi i \ln(\overline{a} - b) + 2 i \iint_D \frac{1}{(\overline{z} - b)(z-a)} dx\wedge dy$$
However I don't know how to perform the double integration on the right. I'm guessing an application of Green's theorem would do the trick but I don't know how to do that with the $\overline{z}$ term.
And actually I have no idea if I'm doing any of this right. I could be grossly misunderstanding something. Any help would be appreciated.