Integral related to a geometry problem In the question Geometry problem involving infinite number of circles I showed that the answer could be obtained by the sum
$$
\sum_{k=0}^{\infty}\int_{B_{k}}
{4 \over \,\left\vert\,1 + \left(\,x + y{\rm i}\,\right)\,\right\vert^{\,4}\,}
\,{\rm d}x\,{\rm d}y\,,
$$
where
$
B_k = \left\{\, z \in \mathbb{C}:\
\left\vert\,z - \left[\frac{1}{2} + \left(k + {1 \over 2}\right)
{\rm i}\right]\,\right\vert\ \leq\ {1 \over 2}\,\right\}
$
is the ball centered at
${1 \over 2} + \left(k + {1 \over 2}\right){\rm i}$ and radius ${1 \over 2}$.
Eventually I solved the problem in a different manner, and from my answer we can deduce that
$$\int_{B_{k}}{4 \over \left\vert\,1 + (x + iy)\,\right\vert^{\,4}}
\,{\rm d}x\,{\rm d}y={16 \pi \over \left(\,4k^{2} + 4k + 9\,\right)^{2}}.$$
Can we somehow get this result directly ?. If so, I could perhaps shorten my answer quite a bit.
 A: You can compute the integral directly using complex coordinates.
Let $z = x + iy$ and $\bar{z} = x-iy$ and notice
$$dx \wedge dy = \frac{d\bar{z} \wedge dz}{2i}$$
The integral $I_k$ over the disk $B_k$ becomes
$$\int_{B_k} \frac{4}{|1 + (x+iy)|^4} dx \wedge dy
= -2i \int_{B_k} \frac{1}{(1+z)^2(1+\bar{z})^2} d\bar{z} \wedge dz\\
= 2i \int_{B_k} d \left( \frac{1}{(1+\bar{z})(1+z)^2} dz \right)
= 2i \int_{\partial B_k} \frac{dz}{(1 + \bar{z})(1+z)^2}
$$
Let $\omega_k = \frac12 + i\left(k + \frac12\right)$. For $z \in \partial B_k$, parametrize $z$ as $\;\omega_k + \frac12 \rho\;$ for $\rho \in S^1$. We have
$$I_k = i \int_{S^1} \frac{d\rho}{(1 + \bar{\omega}_k + \frac{1}{2\rho})(1 + \omega_k + \frac12\rho)^2}
= \frac{i}{1+\bar{\omega}_k}
\int_{S^1} \frac{\rho d\rho}{
\left(\rho + \frac{1}{2(1+\bar{\omega}_k)}\right)(1 + \omega_k + \frac12\rho)^2}
$$
In the integral at RHS, there is only one pole inside the unit circle $S^1$. Namely, $\rho = -\frac{1}{2(1+\bar{\omega}_k)}$. As a result, the integral is equal to $2 \pi i$ times corresponding residue at $-\frac{1}{2(1+\bar{\omega}_k)}$.
$$\begin{align}
I_k 
&= (2\pi i) \frac{i}{1+\bar{\omega}_k}\frac{-\frac{1}{2(1+\bar{\omega}_k)}}{\left(1 + \omega_k - \frac{1}{4(1 + \bar{\omega}_k)}\right)^2}\\
&= \frac{\pi}{\left(|1 + \omega_k|^2 - \frac14\right)^2}
= \frac{\pi}{\left(
\left(
\frac32\right)^2 + \left(k + \frac12\right)^2 - \frac14
\right)^2}
= \frac{16\pi}{(4k^2+4k+9)^2}
\end{align}
$$
