<s>[INST] 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:
$$\int_0^a\frac{a}{(y^2+a^2)\sqrt{y^2+2a^2}}dy=\frac{\pi}{6a}$$
 [/INST] The natural course of action whenever you see $x^2 + y^2$ in a multiple integration problem is to convert to polar coordinates.
Set $x = r\cos\theta$, $y = r\sin\theta$, $0 \leq \theta <2\pi$. Then $dx~dy = r~dr~d\theta$, and the integrand becomes
$$
\frac{r}{(r^2 + a^2)^{3/2}}~dr~d\theta.
$$
This integral doesn't depend on $\theta$, so in fact when you integrate in $r$ it is a one-variable integral, and from single-variable calculus we know that the substitution $u = r^2 + a^2$ will do the trick. The only thing left to do is to find the limits of integration.
The region $0 \leq x \leq a$, $0 \leq y \leq a$ is a square of side length $a$ in the first quadrant of the plane. This can be parametrized by $0 \leq \theta \leq \frac{\pi}{4}$, $0 \leq r \leq \sqrt{a^2 + (a\sin\theta)^2} = a\sqrt{1 + \sin^2 \theta}$. (Draw a picture! This is the meat of the problem.) Therefore the integral becomes
$$
\int_{\theta = 0}^{\pi/4}\int_{r=0}^{a\sqrt{1+\sin^2\theta}} \frac{r}{(r^2 + a^2)^{3/2}}~dr~d\theta.
$$
Can you take it from here?
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