Double Integral $\int\limits_0^a\int\limits_0^a\frac{dx\,dy}{(x^2+y^2+a^2)^\frac32}$ How to solve this integral?
$$\int_0^a\!\!\!\int_0^a\frac{dx\,dy}{(x^2+y^2+a^2)^\frac32}$$
my attempt
$$
\int_0^a\!\!\!\int_0^a\frac{dx \, dy}{(x^2+y^2+a^2)^\frac{3}{2}}=
\int_0^a\!\!\!\int_0^a\frac{dx}{(x^2+\rho^2)^\frac{3}{2}}dy\\
\rho^2=y^2+a^2\\
x=\rho\tan\theta\\
dx=\rho\sec^2\theta \, d\theta\\
x^2+\rho^2=\rho^2\sec^2\theta\\
\int_0^a\!\!\!\int_0^{\arctan\frac{a}{\rho}}\frac{\rho\sec\theta}{\rho^3\sec^3\theta}d\theta \, dy=
\int_0^a\!\!\!\frac{1}{\rho^2}\!\!\!\int_0^{\arctan\frac{a}{\rho}}\cos\theta \, d\theta \, dy=\\
\int_0^a\frac{1}{\rho^2}\sin\theta\bigg|_0^{\arctan\frac{a}{\rho}} d\theta \, dy=
\int_0^a\frac{1}{\rho^2}\frac{x}{\sqrt{x^2+\rho^2}}\bigg|_0^ady=\\
\int_0^a\frac{a}{(y^2+a^2)\sqrt{y^2+2a^2}}dy$$
Update:
$$\int_0^a\frac{a}{(y^2+a^2)\sqrt{y^2+2a^2}}dy=\frac{\pi}{6a}$$

Both the function that you are integrating as the region over which you are integrating it get unchanged if you exchange $x$ with $y$. Therefore, your integral is equal to$$2\int_0^a\int_0^x\frac1{(a^2+x^2+y^2)^{3/2}}\,\mathrm dy\,\mathrm dx.$$You can compute this integral using polar coordinates: $\theta$ can take values in $\left[0,\frac\pi4\right]$ and, for each $\theta$, $r$ can take values in $\left[0,\frac a{\cos\theta}\right]$. And\begin{align}\int_0^{\pi/4}\int_0^{a/\cos(\theta)}\frac r{(a^2+r^2)^{3/2}}\,\mathrm dr\,\mathrm d\theta&=\int_0^{\pi/4}\frac{1-\frac1{\sqrt{\sec ^2(\theta)+1}}}a\,\mathrm d\theta\\&=\frac1a\left(\frac\pi4-\int_0^{\pi/4}\frac1{\sqrt{\sec ^2(\theta)+1}}\,\mathrm d\theta\right)\\&=\frac\pi{12a}.\end{align}Note that the final equality is equivalent to$$\int_0^{\pi/4}\frac1{\sqrt{\sec^2(\theta)+1}}\,\mathrm d\theta=\frac\pi6.$$This can be justified as follows: you do $\theta=\arccos\left(\sqrt x\right)$ and $\mathrm d\theta=-\frac1{2\sqrt{x-x^2}}\,\mathrm dx$. Doing this, you will get\begin{align}\int_0^{\pi/4}\frac1{\sqrt{\sec^2(\theta)+1}}\,\mathrm d\theta&=\int_1^{1/2}-\frac1{2\sqrt{1-x^2}}\\&=\frac12\int_{1/2}^1\frac1{\sqrt{1-x^2}}\,\mathrm dx\\&=\frac12\left(\arcsin\left(1\right)-\arcsin\left(\frac12\right)\right)\\&=\frac12\left(\frac\pi2-\frac\pi6\right)\\&=\frac\pi6.\end{align}
