Double Integral Question on unit square Hints on solving following double integral will be appreciated.
$$\int_0^1 \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \text{d}y \ \text{d}x$$

Just split the integral as
$$ \int_0^1 \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \text{d}y \ \text{d}x
=  \int_0^1 \int_0^1 \frac{x^2}{(x^2+y^2)^2} \text{d}y \ \text{d}x
- \int_0^1 \int_0^1 \frac{y^2}{(x^2+y^2)^2} \text{d}y \ \text{d}x. $$
Now, just use standard integration techniques to evaluate the integrals. For instance 
$$ \int_{0}^{1}\frac{1}{(x^2+y^2)^2}dy=\frac{xy}{x^2 + y^2} + \frac{\arctan(y/x))}{2x}\Big|_{0}^{1} =\dots.  $$
