Integral of functions of several variables The function is defined from $[-1,1]\times[-1,1]$ to $\Bbb R$, given by $f(x,y)=\frac{x^2-y^2}{(x^2+y^2)^2}$ when $(x,y)≠0$ and $f(0,0)=0$.
I could find that the function is not continuous at origin, so not differentiable  and also partial derivatives does not exist at origin.
But how do we evaluate the integral of the function over the given domain. Being function of several variable how do we deal with the point of discontinuity? Can we convert it to polar coordinates and apply residue theorem? Do we have any other methods?
Any help would be appreciated. Thanks in advance.

This is a (somewhat?) standard example of how rearranging an iterated integral makes it converge conditionally to either of two different numbers.
\begin{align}
& \int_0^1 \left( \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \, dy \right) \, dx = +\frac \pi 4. \\[10pt]
& \int_0^1 \left( \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \, dx \right) \, dy = -\frac \pi 4. \\[10pt]
& \iint\limits_{0\,<\,y\,<\,x\,<\,1} \frac{x^2-y^2}{(x^2+y^2)^2} \, d(x,y) = +\infty. \\[10pt]
& \iint\limits_{0\,<\,x\,<\,y\,<\,1} \frac{x^2-y^2}{(x^2+y^2)^2} \, d(x,y) = -\infty. \\[10pt]
& \iint\limits_{\varepsilon\,<\,x\,<\,1 \\ \varepsilon\,<\,y\,<\,1} \frac{x^2-y^2}{(x^2+y^2)^2} \, d(x,y) = 0.
\end{align}
Only when the integrals of the positive and negative parts are both infinite can the values of the two iterated integrals differ.
Let $y = x\tan \theta$, so that $dy = x\sec^2\theta\,d\theta$ and
$x^2 + y^2= x^2\sec^2\theta$, and as $y$ goes from $0$ to $1$
then $\theta$ goes from $0$ to $\arctan(1/x)$. Then
\begin{align*}
\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \, dy
= {} & \int_0^{\arctan(1/x)}
\frac{x^2 - x^2 \tan^2\theta}{(x^2 + x^2\tan^2\theta)^2} \big( x\sec^2\theta\,d\theta\big) \\[10pt]
= {} & \frac 1 x \int_0^{\arctan(1/x)} \frac{1-\tan^2 \theta}{\sec^2 \theta} \, d\theta \\[10pt]
= {} & \frac 1 x \int_0^{\arctan(1/x)} (\cos^2\theta-\sin^2\theta) \, d\theta \\[10pt]
= {} & \frac 1 x \int_0^{\arctan(1/x)} \cos(2\theta) \, d\theta \\[10pt]
= {} & \frac 1 {2x} \sin\left(2\arctan \frac 1 x\right) \\[10pt]
= {} & \frac 1 x \sin\left(\arctan \frac 1 x \right) \cos\left( \arctan \frac 1 x \right) \\[10pt]
= {} & \frac 1 x \cdot \frac 1 {\sqrt{1+x^2}} \cdot \frac x {\sqrt{1+x^2}} = \frac 1 {1+x^2}. \\[10pt]
\text{And then}
& \int_0^1 \frac{dx}{1+x^2} = \frac \pi 4.
\end{align*}
