Why is $\, \int_0^1 \{ \int_0^1 \frac{x-y}{(x+y)^3} \, dy \} \, dx \, \neq \,\int_0^1 \{ \int_0^1 \frac{x-y}{(x+y)^3} \, dx \} \, dy \,$? As far I know: Double Integrals of a function depend only on (i)Region of Integration and (ii)Function, and not on its order of integration.
In this case :
(i)Region of integration is a square (ABCD with AB=BC=CD=DA=$1$ units) where one of its vertices (A) lies on the origin and the opposite vertex is at C=($1,1$);
(ii) Function: $ f(x,y)=\frac{x-y}{(x+y)^3}$
Then, why is
$$\int_0^1 \{ \int_0^1 \frac{x-y}{(x+y)^3} \, dy \} \, dx \, \neq \,\int_0^1 \{ \int_0^1 \frac{x-y}{(x+y)^3} \, dx \} \, dy \,\,\,\,?$$
i.e.,
$$\int_0^1 \{ \int_0^1 \frac{x-y}{(x+y)^3} \, dy \} \, dx \, =0.5 \,\, \& \,\,\int_0^1 \{ \int_0^1 \frac{x-y}{(x+y)^3} \, dx \} \, dy \,=-0.5 $$
Have I missed any concepts? Please help...
 A: It is clear that the iterated integrals have opposite sign (seen simply by interchanging labels $x \iff y$.) Unless both are zero, which is not the case, they are unequal.
Your computed values are correct and are obtained easily by noticing that
$$\frac{x-y}{(x+y)^3} = -\frac{\partial}{\partial x} \left( \frac{x}{(x+y)^2} \right)$$
When the integrand is nonnegative or absolutely integrable, then Tonelli's or Fubini's theorem, respectively,  guarantees that the iterated integrals are equal, i.e. the order of integration may be switched.  
In this case both conditions are not met. In some cases the order of integration can be switched without meeting these requirements (by lucky cancellation) but there is no guarantee -- and this integral is a good example.
To see why Fubini's theorem does not apply, note that transforming to polar coordinates we have for $0 < \delta < 1$,
$$\begin{align}\int_{[0,1]^2} \left|\frac{x-y}{(x+y)^3} \right| &\geqslant \int_0^{\pi/2}\int_{\delta}^1 \frac{r|\cos \theta - \sin \theta|}{r^3|\cos \theta + \sin \theta|^3} r \, dr\, d\theta \\ &= \int_0^{\pi/2}\frac{|\cos \theta - \sin \theta|}{|\cos \theta + \sin \theta|^3}  \,d\theta \int_{\delta}^1 \frac{dr}{r} \\ &= - \log \delta\int_0^{\pi/2}\frac{|\cos \theta - \sin \theta|}{|\cos \theta + \sin \theta|^3}  \,d\theta \end{align}$$
and the RHS tends to $+\infty$ as $\delta \to 0$. 
(Splitting the integral on the second line is permissible because the integrand is continuous on $[\delta,1] \times [0, \pi/2]$.)
