How do I prove that $\int_0^1 \int_0^1 {{2+x^2-y^2}\over {2-x^2-y^2}} \, dx \, dy=2G?$ Let $G$ the Catalan's constant, prove that

$$\int_0^1 \int_0^1 {{2+x^2-y^2}\over {2-x^2-y^2}} \, dx \, dy=2G$$

I not sure what do, anyway I try to integrate with respect to x first
$$\int_0^1 {2-y^2+x^2\over 2-y^2-x^2}\,dx$$
Make a substitution $x^2=2+y^2$
$$-\int_0^1 {2\over xy} \, dy$$
$$-\int_0^1 {2\over y\sqrt{2+y^2}} \, dy$$
I can't find a standard integral for this, some help please.
 A: First we can observe, by symmetry, that $$\int_{0}^{1}\int_{0}^{1}\frac{x^{2}-y^{2}}{2-x^{2}-y^{2}}dxdy=\int_{0}^{1}\int_{0}^{1}\frac{x^{2}}{2-x^{2}-y^{2}}dxdy-\int_{0}^{1}\int_{0}^{1}\frac{y^{2}}{2-x^{2}-y^{2}}dxdy=0
 $$ hence $$I=\int_{0}^{1}\int_{0}^{1}\frac{2+x^{2}-y^{2}}{2-x^{2}-y^{2}}dxdy=2\int_{0}^{1}\int_{0}^{1}\frac{1}{2-x^{2}-y^{2}}dxdy.
 $$ Now observe, again by symmetry, that the contributions with $x\leq y
  $ and $y\leq x
 $ are the same, so $$I=4\int_{0}^{1}\int_{0}^{x}\frac{1}{2-x^{2}-y^{2}}dydx.
 $$ Now taking the polar coordinates we have $$I=2\int_{0}^{\pi/4}\int_{0}^{\sec\left(\theta\right)}\frac{2\rho}{2-\rho^{2}}d\rho d\theta
 $$ $$=2\int_{0}^{\pi/4}\log\left(\frac{2}{2-\sec^{2}\left(\theta\right)}\right)d\theta
 $$ but $$\frac{2}{2-\sec^{2}\left(\theta\right)}=\frac{\tan\left(2\theta\right)}{\tan\left(\theta\right)}
 $$ hence $$I=2\int_{0}^{\pi/4}\log\left(\tan\left(2\theta\right)\right)d\theta-2\int_{0}^{\pi/4}\log\left(\tan\left(\theta\right)\right)d\theta
 $$ and now it is sufficient to note that $$2\int_{0}^{\pi/4}\log\left(\tan\left(2\theta\right)\right)d\theta=\int_{0}^{\pi/2}\log\left(\tan\left(u\right)\right)du=0
 $$ since $$\int_{0}^{\pi/2}\log\left(\sin\left(u\right)\right)du=\int_{0}^{\pi/2}\log\left(\sin\left(\frac{\pi}{2}-u\right)\right)du=\int_{0}^{\pi/2}\log\left(\cos\left(u\right)\right)du
 $$ (and, for personal culture, $\int_{0}^{\pi/2}\log\left(\sin\left(u\right)\right)du=-\frac{\pi}{2}\log\left(2\right)
 $) so we have done $$I=-2\int_{0}^{\pi/4}\log\left(\tan\left(\theta\right)\right)d\theta=\color{red}{2G}$$ as wanted.
