Integration in a disk in 2D plane I want to ask how to integrate $\int_{B_r(0)} \dfrac{1}{\sqrt{|x|}|x-a|^2}dx$ where $x \in \mathbb{R}^2$, does it has a explicit solution? or some estimate?
Thank you!!
 A: Let $I$ be the integral given by
$$I=\int_{-\pi}^\pi \int_0^r \frac{\sqrt{\rho}}{\rho^2-2|\vec a|\rho \cos(\phi)+|\vec a|^2}\,d\rho \,d\phi$$
Now, we note that the integral diverges for $|\vec a|<r$.  Therefore, we assume that $|\vec a|>r$.
We can evaluate the iterated integral over $\phi$ by either using the Tangent Half-Angle Substitution or contour integration.  Proceeding, we find 
$$\begin{align}
\int_{-\pi}^\pi  \frac{1}{\rho^2-2|\vec a|\rho \cos(\phi)+|\vec a|^2} \,d\phi&=2\int_{0}^\pi  \frac{1}{\rho^2-2|\vec a|\rho \cos(\phi)+|\vec a|^2} \,d\phi\\\\
&=4\int_0^\infty \frac{1}{(\rho^2+|\vec a|^2)(1+t^2)-2|\vec a|\rho (1-t^2)}\,dt\\\\
&=4\int_0^\infty \frac{1}{(|\vec a|+\rho)^2t^2+(|\vec a|-\rho)^2}\,dt\\\\
&=\frac{2\pi}{|\vec a|^2-\rho^2}
\end{align}$$
Then, the integral of interest, $I$, reduces to 
$$\begin{align}
I&=2\pi \int_0^r \frac{\sqrt{\rho}}{|\vec a|^2-\rho^2}\,d\rho\\\\
&=\bbox[5px,border:2px solid #C0A000]{\frac{2\pi}{\sqrt{|\vec a|}} \left(\text{arctanh}\left(\sqrt{\frac{r}{|\vec a|}}\right)-\arctan\left(\sqrt{\frac{r}{|\vec a|}}\right)\right) } \tag 2
\end{align}$$
where we arrived at $(2)$ by enforcing the substitution $\rho \to \rho^2$, and following that with partial fraction expansion.
