aruwin said:
Ok, I have decomposed the fractions successfully and I got the answer -1/2 which is wrong. The answer given to me is undefined (infinity). Is there any explanation to this?
Ah, yes, I didn't notice it before, but this is a tricky question. If you follow my previous hint then you get $$\begin{aligned} \int_0^1 \int_0^1 \frac{x-y}{(x+y)^3}\,dx\,dy &= \int_0^1 \int_0^1 \bigl((x+y)^{-2} - 2y(x+y)^{-3}\bigr)\,dx\,dy \\ &= \int_0^1 \Bigl[-(x+y)^{-1} + y(x+y)^{-2}\Bigr]_{x=0}^1dy \\ &= \int_0^1 \bigl(-(y+1)^{-1} + \rlap{\color{red}/}y^{-1} + y(y+1)^{-2} - \rlap{\color{red}/}y^{-1}\bigr)\,dy. \end{aligned}$$
If you now cancel those $y^{-1}$s and integrate, then you get the answer -1/2 (which is wrong). However, some alarm bells should have started to sound when you did the cancellation, because the y-integral goes from 0 to 1, and the function $y^{-1}$ becomes infinite at the left end of that interval.
Another cause for concern is that if you use the same method to do the y-integral first, before the x-integral, then you would end up with the result +1/2 rather than -1/2. (You don't actually have to do the calculation to see that. Just observe that the function $(x-y)/(x+y)^3$ is anti-symmetric: it changes sign when you interchange $x$ and $y$.)
What all this shows is that this integral is an improper integral. The function becomes infinite at the origin, which is at one corner of the domain of integration. As with functions of a single variable, you have to be very cautious with such integrals, and ensure that the improper integral converges. In this case, it does not.