Double Integral I need help calculating these double integrals (in order to show they are not equal):
$$\int_0^1\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} dy\,dx \ne \int_0^1\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} dx\,dy$$
 A: Consider the double integral over the rectangle $[0,1]\times[0,1]$
$$\iint\limits_{[0,1]\times[0,1]} \! \frac{x^2-y^2}{(x^2+y^2)^2} \, \mathrm{d}y \, \mathrm{d}x.$$ 
Partial fraction decomposition yields
$$=\iint\limits_{[0,1]\times[0,1]} \! \frac{2x^2}{(x^2+y^2)^2} - \frac{1}{x^2+y^2} \, \mathrm{d}y \, \mathrm{d}x,$$
$$=\int\limits_{[0,1]} \! \frac{y}{x^2+y^2} \Bigg|_{y=0}^{y=1} \, \mathrm{d}x = \int\limits_{[0,1]} \! \frac{\mathrm{d}x}{x^2+1} = \frac{\pi}{4}.$$
We then consider the double integral over the rectangle $[0,1]\times[0,1]$
$$\iint\limits_{[0,1]\times[0,1]} \! \frac{x^2-y^2}{(x^2+y^2)^2} \, \mathrm{d}x \, \mathrm{d}y.$$
$$=\iint\limits_{[0,1]\times[0,1]} \! \frac{2x^2}{(x^2+y^2)^2} - \frac{1}{x^2+y^2} \, \mathrm{d}x \, \mathrm{d}y,$$
$$=-\int\limits_{[0,1]} \! \frac{x}{x^2+y^2} \Bigg|_{x=0}^{x=1} \, \mathrm{d}y = -\int\limits_{[0,1]} \! \frac{\mathrm{d}y}{y^2+1} = -\frac{\pi}{4}.$$
Therefore,
$$\iint\limits_{[0,1]\times[0,1]} \! \frac{x^2-y^2}{(x^2+y^2)^2} \, \mathrm{d}y \, \mathrm{d}x \neq\iint\limits_{[0,1]\times[0,1]} \! \frac{x^2-y^2}{(x^2+y^2)^2} \, \mathrm{d}x \, \mathrm{d}y$$
