Evaluate $\iint_{[0,1]^2}\frac{dxdy}{(1+x^2+y^2)^{3/2}}$ 
Evaluate: $$I=\iint_{[0,1]^2}\frac{dxdy}{(1+x^2+y^2)^{3/2}}.$$

Attempt. Working on substitution $x=\sqrt{1+y^2}\,\sinh u$ we get
$$I=\int\limits_{0}^{1}\frac{x}{(1+y^2)\sqrt{x^2+y^2+1}}\Bigg|_{0}^{1}\,dy=\int\limits_{0}^{1}\frac{dy}{(1+y^2)\sqrt{2+y^2}}$$ and then working on $y=\sqrt{2}\,\tan u$ we get:
$$I=\arctan\left(\frac{y}{\sqrt{y^2+2}}\right)\bigg|_{0}^{1}=\arctan\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6}.$$
On the other hand, polar coordinates give:
$$I=\int\limits_{0}^{\pi/4}\int\limits_{0}^{\frac{1}{\cos \phi}}\frac{r}{(1+r^2)^{3/2}}dr d\phi+\int\limits_{\pi/4}^{\pi/2}\int\limits_{0}^{\frac{1}{\sin \phi}}\frac{r}{(1+r^2)^{3/2}}dr d\phi$$
which gets us to:
$$I=\int\limits_{0}^{\pi/4}\left(1-\frac{\cos\phi}{\sqrt{1+\cos^2\phi}}\right)d\phi+\int\limits_{\pi/4}^{\pi/2}\left(1-\frac{\sin\phi}{\sqrt{1+\sin^2\phi}}\right)d\phi$$
and the computations seem difficult to handle.
Is there an easier way to approach the calculation of this integral?
Thank you in advance.
 A: First, note that the two integrals in the last line are equal. Just substitute $\phi=\frac{\pi}{2}-\varphi$ and see. Therefore
$$I=2\int_{0}^{\frac{\pi}{4}}\left(1-\frac{\cos\phi}{\sqrt{2-\sin^{2}\phi}}\right){\rm d\phi}=\left[\begin{matrix}\sin\phi=\sqrt{2}\sin\theta\\\cos\phi{\rm d}\phi=\sqrt{2}\cos\theta{\rm d}\theta\end{matrix}\right]=$$
$$=\frac{\pi}{2}-2\int_{0}^{\frac{\pi}{6}}\frac{\cos\theta}{\sqrt{1-\sin^{2}\theta}}{\rm d}\theta=\frac{\pi}{2}-\frac{\pi}{3}=\frac{\pi}{6}$$
A: Here is a solution that takes the issue right at its beginning. Your idea to turn to polar coordinates and divide the integration domain into two triangles $OAB$ and $OBC$ (with $A(1,0),B(1,1)$ and $C(0,1)$) is good. You should do it at once.  One gets :
$$I=\iint_{[0,1]^2}\frac{dxdy}{(1+x^2+y^2)^{3/2}}=I_1+I_2$$
where 
$$I_1=\int_{\theta=0}^{\theta=\pi/4}\int_{r=0}^{r=1/\cos{\theta}}\frac{1}{(1+r^2)^{3/2}}\color{red}{r}drd\theta=$$
and $I_2$ is in fact equal to $I_1$ for an evident symmetry reason. (please note that $\color{red}{r}$ is the jacobian of this change of variables). Thus :
$$I=2I_1=2\int_{\theta=0}^{\theta=\pi/4}[-\frac{1}{(1+r^2)^{1/2}}]_{r=0}^{r=1/\cos(\theta)} d\theta$$
giving :
$$I=2\int_{\theta=0}^{\theta=\pi/4}(1-\frac{1}{(1+(1/\cos(\theta)^2)^{1/2}})d\theta=2\left(\frac{\pi}{4}-\int_{\theta=0}^{\theta=\pi/4}\frac{\cos(\theta)}{\sqrt{1+\cos(\theta)^2}}d\theta\right)$$
where the remaining integral, transformed into :
$$\int_{\theta=0}^{\theta=\pi/4}\frac{\cos(\theta)}{\sqrt{2 -\sin(\theta)^2}}d\theta$$
is readily integrated using antiderivative :
$$\sin^{-1}\left(\tfrac{1}{\sqrt{2}}\sin(\theta)\right)$$
giving finally :
$$I=2(\frac{\pi}{4}-\frac{\pi}{6})=\frac{\pi}{6}$$
