Finding $\iint_D\frac{x^2-y^2}{\sqrt{x^2+y^2}}\,\mathrm dx\mathrm dy$ where $D=\{(x,y)\in\mathbb{R}^2:\ 0\leq y\leq x\,,xy\leq 1\leq x+y-1\}$ How can I evaluate this double integral through a domain transformation?

$$\iint_D\frac{x^2-y^2}{\sqrt{x^2+y^2}}\,\mathrm dx\mathrm dy$$
  where $$D=\{(x,y)\in\mathbb{R}^2:\ 0\leq y\leq x\,, xy\leq 1\leq x+y-1\}$$

The region $D$ is the red one in this picture:

I have tried to convert $D$ to polar coordinates but then, it come up the following integral:
$$\frac{1}{3}\int_0^{\frac{\pi}{4}}\cos 2t\left(\frac{1}{\sqrt{(\sin t\cos t)^3}}-\frac{8}{(\sin t+\cos t)^3}\right)dt$$
which I got stuck into.
Does someone any further ideas?
 A: I have found that is possible to solve this integral by the following transformation:
$$\begin{cases}
u=x+y\\
v=x-y\end{cases}\ \Rightarrow\
\begin{cases}
x=\frac{u+v}{2}\\
y=\frac{u-v}{2}\end{cases}$$
The integral does not converge:
$$\begin{eqnarray}
\iint_D\frac{x^2-y^2}{\sqrt{x^2+y^2}}\ dxdy &=& \int_2^{+\infty}du\int_{\sqrt{u^2-4}}^u\frac{\frac{(u+v)^2}{4}-\frac{(u-v)^2}{4}}{\sqrt{\frac{(u+v)^2}{4}+\frac{(u-v)^2}{4}}}\cdot \left(-\frac{1}{2}\right)\ dv=\\
&=&-\frac{\sqrt{2}}{2} \int_2^{+\infty}du\int_{\sqrt{u^2-4}}^u\frac{v}{\sqrt{u^2+v^2}}\ dv =\\
&=&-\frac{\sqrt{2}}{4} \int_2^{+\infty}u\ du\int_{\sqrt{u^2-4}}^u 2v(u^2+v^2)^{-1/2}\ dv=\\
&=&-\frac{\sqrt{2}}{2} \int_2^{+\infty}u\cdot [\sqrt{u^2+v^2}]_{\sqrt{u^2-4}}^u\ du =\\
&=&-\int_2^{+\infty}u^2-u\sqrt{u^2-2}\ du=\\
&=&-\int_2^{+\infty}u^2\ du +\frac{1}{2}\int_2^{+\infty}2u(u^2-2)^{1/2}\ du=\\
&=&\lim\limits_{c\to +\infty}\left \{\left[-\frac{u^3}{3}\right]_2^c+\frac{1}{3}\left[\sqrt{(u^2-2)^3}\right]_2^c\right \}=\\
&=&\lim\limits_{c\to +\infty}\left(-\frac{c^3}{3}+\frac{8}{3}+\frac{1}{3}\sqrt{(c^2-2)^3}-\frac{2\sqrt{2}}{3}\right)=\\  
&=&\frac{1}{3}\lim\limits_{c\to +\infty}(\sqrt{(c^2-2)^3}-c^3+8-2\sqrt{2})=-\infty\end{eqnarray}$$
