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

Since the integrand has a non-integrable singularity at the origin, we must evaluate it by iterated integration. So let's consider the inner integral first:
$$\begin{align}
\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2}\, dy &= \frac{1}{x}\int_0^{1/x} \frac{1-t^2}{(1+t^2)^2}\, dt \qquad\qquad\qquad\qquad\qquad (y = x\cdot t)\\
&= \frac{1}{2x} \int_0^{2\arctan(1/x)} \frac{1 - \tan^2 (\varphi/2)}{(1+\tan^2(\varphi/2))^2}(1+\tan^2(\varphi/2))\, d\varphi\quad (t = \tan (\varphi/2))\\
&= \frac{1}{2x} \int_0^{2\arctan(1/x)} \frac{\cos^2(\varphi/2) - \sin^2(\varphi/2)}{\cos^2(\varphi/2)+\sin^2(\varphi/2)}\, d\varphi\\
&= \frac{1}{2x} \int_0^{2\arctan(1/x)} \cos\varphi\,d\varphi\\
&= \frac{1}{2x} \sin \left(2\arctan(1/x)\right)\\
&= \frac{1}{x} \sin \left(\arctan(1/x)\right)\cos\left(\arctan(1/x)\right)\\
&= \frac{1}{x} \frac{1/x}{\sqrt{1+1/x^2}}\frac{1}{\sqrt{1+1/x^2}}\\
&= \frac{1}{1+x^2}
\end{align}$$
From the on, it's very simple,
$$\int_0^1 \frac{dx}{1+x^2} = \arctan 1 = \frac{\pi}{4}.$$
Note that the other order of integration,
$$\int_0^1\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2}\, dx\,dy$$
leads to the value $-\frac{\pi}{4}$.
