While reading a paper I stumbled across the integral $$ \iint_{\mathbb{R}^2} \ln \left(\frac{1}{|x-y|}\right) \frac{4}{(1+|y|^2)^2} d^2y\,, $$ where $x \in \mathbb{R}^2$ is fixed. I think the paper implicitly claims that this is equal to $-2 \pi \ln (1+|x|^2)$.
I'm not sure if this is a hard integral at all, but considering the amount of proficient integrators on this site, I thought that I should post this question here.
To provide some context, the term $\ln \frac{1}{|x-y|}$ comes from the Laplacian's Green function (it is the leading term near $x$) and the term $\frac{4}{(1+|y|^2)^2}$ is of course the "density" of the stereographic projection metric.
Please note the result $$\int_0^{2\pi}\!d\phi\, \ln(a^2- 2a b \cos(\phi) +b^2) = 4\pi \ln[\max(|a|,|b|)] \tag{1}$$ which can be proven by different methods (e.g., by taking the derivative with respect to $a$ or $b$).
Your integral is given by (using polar coordinates $(r,\phi)$) $$I= \iint_{\mathbb{R}^2}\!d^2y \ln\left(\frac{1}{|x-y|}\right) \frac{4}{(1+|y|^2)^2} = -\int_0^\infty \!r dr\int_{0}^{2\pi} \!d\phi \, \ln(|x|^2 - 2 |x| r\cos \phi + r^2) \frac{2}{(1+r^2)^2} .$$
With the result (1), we obtain $$ I = - 4\pi\int_0^\infty\!dr \frac{2 r}{(1+r^2)^2} \ln[\max(|x|,|r|)] = -4\pi \ln |x| \int_0^{|x|}\!dr\,\frac{2r}{(1+r^2)^2} -4\pi \int_{|x|}^\infty\!dr\,\frac{2 r \ln r}{(1+r^2)^2} \\ = -\frac{4\pi |x|^2 \ln |x|}{1+|x|^2} - 2\pi\left[ \frac{2r^2 \ln r}{1+r^2} -\ln (1+r^2) \right]_{r=|x|}^\infty = -2\pi \ln(1+|x|^2)$$