Evaluate $\iint (\frac{(x-y)}{x+y})^4$ over the triangular region bounded by $x+y=1$, $x$-axis ,$y$-axis 
Evaluate $\iint (\frac{(x-y)}{x+y})^4$ over the triangular region bounded by $x+y=1$, $x$-axis, $y$-axis.

My attempt: I tried using the change of variable concept:
Let $u=x-y$ and $v=x+y$ ,$|J|= \frac{1}{2}$
Then $x=\frac{u+v}{2}$ and $y=\frac{u-v}{2}$
Now we have to find the limit which is where I am stuck.
We were told in class that we could plug in the points and get the equations:
In this case the points are $(0,0),(0,1),(1,0)$.
So if we put the first point we get $u=0,v=0$ or $u=-v,u=v$
If we put in the second point we get
$u=-1,v=1$, $u=-v,u=v+2$
Similarly
$u=1,v=-1$,$u=v,u=v-2$
How do I proceed after this? This is making me even more confused on what values to take and what values not to take.can anyone help me out here with an easy explanation?

No change of variable seems necessary.I'll use Sage to explain.
Define the function :
$$f(x,y)=\left(\frac{x-y}{x+y}\right)^4$$
sage: f(x,y)=((x-y)/(x+y))^4

Find an antiderivative of this function with respect to $y$ :
sage: A1=integrate(f(x, y), y) ; A1
-8*x*log(x + y) + y - 8/3*(5*x^4 + 12*x^3*y + 9*x^2*y^2)/(x^3 + 3*x^2*y + 3*x*y^2 + y^3)

$$\int\left(\frac{x-y}{x+y}\right)^4\,dy\,=\,-8x\log(x + y) + y - \frac{8}{3}\frac{5x^4 + 12x^3y + 9x^2y^2}{x^3 + 3x^2y + 3xy2 + y^3}$$
This function is defined and continuous on the square $(0,\ 1],(0,\ 1]$ except in $(0,\ 0)$. Its limits give us the defined integral $\displaystyle\int_0^{1-x}f(x,y)\,dy$ :
sage: D1 = A1.limit(y=1-x)-A1.limit(y=0) ; D1
-16/3*x^4 + 16*x^3 - 24*x^2 + 8*x*log(x) + 37/3*x + 1

$$\int_0^{1-x}\left(\frac{x-y}{x+y}\right)^4\,dy\,=\,-\frac{16}{3}x^4 + 16x^3 - 24x^2 + 8x\log(x) + \frac{37}{3}x + 1$$
which is a function of $x$, defined and continuous on $\mathbb{R}^{*+}$, which has an antiderivative :
sage: A2 = integrate(D1, x) ; A2
-16/15*x^5 + 4*x^4 - 8*x^3 + 4*x^2*log(x) + 25/6*x^2 + x

$$\int\frac{16}{3}x^4 + 16x^3 - 24x^2 + 8x\log(x) + \frac{37}{3}x + 1\,dx\,=\,-\frac{16}{15}x^5 + 4x^4 - 8x^3 + 4x^2\log(x) + \frac{25}{6}x^2 + x$$
continuous on $\mathbb{R}$, whose limits give us the sought integral :
sage: D2 = A2.limit(x=1) - A2.limit(x=0) ; D2
1/10

BTW :
sage: integrate(integrate(f(x, y), y, 0, 1-x, algorithm="sympy"), x, 0, 1)
1/10

(The algorithm="sympy" parameter avoids Sage's default integrator (Maxima), whic is nosy about the value of $x$).
