Fast way to integrate $\frac{x^2-y^2}{(x^2+y^2)^2} dx \,dy$ in unit square I am looking for a fast way to integrate $$ \int_0^1 \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} dx \,dy$$ using standard techniques ( no complex analysis and no functional analysis). 
I am aware that wolframalpha spits out a solution, but this one is quite long, I assume that there is a faster way to do this. The result will be $\frac{\pm \pi}{2}$(depending on which integral you do first). By the way: Please be aware of the fact, that fubini's theorem does not hold in this case.

Start with the inside integral:
$$
\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} dx
$$
Let $x=y\tan\theta$, so that $dx = y\sec^2\theta\,d\theta$ and $x^2+y^2 = y^2\tan^2\theta+y^2 = y^2\sec^2\theta$.  As $x$ goes from $0$ to $1$, $\theta$ goes from $0$ to $\arctan(1/y)$, so the integral becomes
\begin{align}
& \phantom{={}}\int_0^{\arctan(1/y)} \frac{y^2\tan^2\theta-y^2}{(y^2\tan^2\theta+y^2)^2} y\sec^2\theta\,d\theta \\[12pt]
& = \frac1y\int_0^{\arctan(1/y)} \frac{\tan^2\theta-1}{\sec^4\theta}\sec^2\theta\,d\theta \\[12pt]
& =\frac1y\int_0^{\arctan(1/y)}(\sin^2\theta-\cos^2\theta)\,d\theta \\[12pt]
& = \frac1y\int_0^{\arctan(1/y)} -\cos(2\theta)\,d\theta \\[12pt]
& = \frac1y\left[\frac{-\sin(2\theta)}{2}\right]_{\theta=0}^{\theta=\arctan(1/y)} = \frac1y\left[-\sin\theta\cos\theta\right]_{\theta=0}^{\theta=\arctan(1/y)}. \tag 1
\end{align}
Since tangent${}={}$opposite$/$adjacent, draw a right triangle in which the "opposite" side is $1$ and the "adjacent" side is $y$.  By the Pythagorean theorem, the hypotenuse is $\sqrt{1+y^2}$, so the sine is opposite$/$hypotenuse $=\dfrac{1}{\sqrt{1+y^2}}$ and the cosine is adjacent$/$hypotenuse $=\dfrac{y}{\sqrt{1+y^2}}$.  Hence $(1)$ becomes
$$
\frac{-1}{1+y^2}.
$$
It's easy to integrate that from $0$ to $1$.
