How to solve $\int_0^1dx\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dy$ The original question is:

Prove that:$$\begin{aligned}\\
\int_0^1dx\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dy\neq\int_0^1dy&\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dx\\
\end{aligned}\\$$

But I can't evaluate the integral $$\int_0^1dx\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dy$$
At first, I assumed $x^2+y^2=z^2$. But, it is so complicated. Then, I assumed $x=r\cos\theta$ and $y=r\sin\theta$. But, I can't calculate the limits. Solving the equations I got three values of $\theta$ i.e. $\theta=0$, $\theta=\frac{\pi}{4}$ and $\theta=\frac{\pi}{2}$. I am just confused. Please help.

Hint:
$$\int_0^1\frac {x^2-y^2}{(x^2+y^2)^2}\, dy = \int_0^1\frac{\partial}{\partial y}  \left(\frac{y}{x^2 + y^2}\right)\, dy =  \frac{1}{1+x^2} $$
