How to compute this double integral I'm trying to show that $\int_0^1\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}dxdy \neq \int_0^1\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}dydx$ by computing these integrals directly. 
I tried using polar coordinates with no success as the bounds of integration caused problems. 
I also tried the substitution $x=ytan\theta$ but ended up getting something of the form $\infty-\infty$. 
Can anyone offer a hint as to how I can compute these directly please??? 
Thanks in advance!

Put $$f(x,y) = \arctan\left( \tfrac x y \right), \,\,\,\,\,\, (x,y) \in (0,1).$$ We see $$\frac{\partial f}{\partial y} = \frac{1}{1+\left(\tfrac x y \right)^2}\left( - \frac x {y^2} \right) = -\frac{x}{x^2 + y^2}.$$ Then $$\frac{\partial }{\partial x}\frac{\partial f}{\partial y} = - \frac{1}{x^2 + y^2} + \frac{x}{(x^2 + y^2)^2}(2x) = \frac{x^2 - y^2}{(x^2+y^2)^2}.$$ Next put $$g(x,y) = -\arctan\left( \tfrac y x \right), \,\,\,\,\,\, (x,y) \in (0,1).$$ Then $$\frac{\partial g}{\partial x} = \frac{1}{1+ \left(-\tfrac{y}{x}\right)^2} \left(\frac{y}{x^2} \right) = \frac{y}{x^2 + y^2}$$ so $$\frac{\partial}{\partial y}\frac{\partial g}{\partial x} = \frac{1}{x^2+y^2} - \frac{y}{(x^2+y^2)^2}(2y) = \frac{x^2 - y^2}{(x^2+y^2)^2}.$$ Then \begin{align*} \int^1_0 \int^1_0 \frac{x^2 - y^2}{(x^2+y^2)^2} dx dy &= \int^1_0 \int^1_0 \frac{\partial }{\partial x}\frac{\partial f}{\partial y} dx dy \\ 
&= \int^1_0 \frac{\partial f}{ \partial y}(1,y) - \frac{\partial f}{ \partial y}(0,y) dy \\
&=- \int^1_0 \frac{1}{1+y^2} dy = - \arctan(1) = - \frac \pi 4
\end{align*} and \begin{align*} \int^1_0 \int^1_0 \frac{x^2 - y^2}{(x^2+y^2)^2} dy dx &= \int^1_0 \int^1_0 \frac{\partial }{\partial y}\frac{\partial g}{\partial x} dy dx \\ 
&= \int^1_0 \frac{\partial g}{ \partial x}(x,1) - \frac{\partial g}{ \partial x}(x,0) dy \\
&= \int^1_0 \frac{1}{1+x^2} dx =  \arctan(1) =  \frac \pi 4.
\end{align*}
