¿Cómo resolver esta integral?
$$\int_0^a\!\!\!\int_0^a\frac{dx\,dy}{(x^2+y^2+a^2)^\frac32}$$
mi intento
$$ \int_0^a\!\!\!\int_0^a\frac{dx \, dy}{(x^2+y^2+a^2)^\frac{3}{2}}= \int_0^a\!\!\!\int_0^a\frac{dx}{(x^2+\rho^2)^\frac{3}{2}}dy\\ \rho^2=y^2+a^2\\ x=\rho\tan\theta\\ dx=\rho\sec^2\theta \, d\theta\\ x^2+\rho^2=\rho^2\sec^2\theta\\ \int_0^a\!\!\!\int_0^{\arctan\frac{a}{\rho}}\frac{\rho\sec\theta}{\rho^3\sec^3\theta}d\theta \, dy= \int_0^a\!\!\!\frac{1}{\rho^2}\!\!\!\int_0^{\arctan\frac{a}{\rho}}\cos\theta \, d\theta \, dy=\\ \int_0^a\frac{1}{\rho^2}\sin\theta\bigg|_0^{\arctan\frac{a}{\rho}} d\theta \, dy= \int_0^a\frac{1}{\rho^2}\frac{x}{\sqrt{x^2+\rho^2}}\bigg|_0^ady=\\ \int_0^a\frac{a}{(y^2+a^2)\sqrt{y^2+2a^2}}dy$$
Actualización:
$$\int_0^a\frac{a}{(y^2+a^2)\sqrt{y^2+2a^2}}dy=\frac{\pi}{6a}$$