Obtain different answers if we change the order of integration In Wolfram alpha, it gives us 
$$\int_0^1 \int_0^1 \frac{x-y}{(x+y)^3} dydx=\frac{1}{2}$$
$$\int_0^1 \int_0^1 \frac{x-y}{(x+y)^3} dxdy=-\frac{1}{2}$$
Why do we have two different answers if we perform the order of integration differently? 
I tried to look up the Fubini's theorem condition, but to no avail. Because the integral above satisfies Fubini's theorem. 
Did I miss up something?
 A: The integral does not satisfy Fubini's theorem. In particular see this part (similar to here).
The article states:

One may switch the order of integration if the double integral yields a finite answer when the integrand is replaced by its absolute value.

However,
$$\int_0^1 \int_0^1 \left|\frac{x-y}{(x+y)^3}\right|\,dx\,dy$$
is not finite, so the theorem does not apply.

To see the above integral is not finite,
\begin{eqnarray}
\int_0^1 \int_0^1 \left|\frac{x-y}{(x+y)^3}\right|\,dx\,dy
&=& \int_0^1 \left(\int_0^y \frac{y-x}{(x+y)^3}dx+\int_y^1 \frac{x-y}{(x+y)^3}dx\right)dy\\
&\ge& \int_0^1 \int_0^y \frac{y-x}{(x+y)^3}dx\,dy = \int_0^1 \int_y^{2y} \frac{2y-u}{u^3}du\,dy\\
&=& \int_0^1 \left[\frac{u-y}{u^2}\right]_{u=y}^{2y}\,dy = \int_0^1 \left[\frac{x}{(x+y)^2}\right]_{x=0}^y\,dy\\
&=& \int_0^1 \frac{1}{4y}\,dy = \frac14 \bigl[\ln(y)\bigr]_{y=0}^1\\
&=& \frac14 \left(\ln(1) - \lim_{t\to 0} \ln(t)\right) = \frac14 \bigl(0 - (-\infty)\bigr)\\
&=& \infty
\end{eqnarray}
where $u = x+y$ was the substitution used in the middle.
