<s>[INST] 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.
 [/INST] Hint:
$$\begin{align}\int\dfrac{x^2 - y^2}{\left(x^2 + y^2\right)^2}\,\mathrm dy&\equiv \int\dfrac{2x^2}{\left(x^2 + y^2\right)^2}  - \dfrac{x^2 + y^2}{\left(x^2 + y^2\right)^2}\,\mathrm dy\\ &= 2x^2\int\dfrac1{\left(x^2 + y^2\right)^2}\,\mathrm dy - \int\dfrac1{x^2 + y^2}\,\mathrm dy\end{align}$$
The first integral can be solved with the help of a reduction formula. The second integral is straightforward to solve (substitute $t = \dfrac yx$ if you get stuck).
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