I need to solve the following integral using change of variables: $$\int\int_D\frac{\sqrt[3]{y-x}}{1+x+y}dA$$ where D is the triangle with vertices $(0,0)$, $(1,0)$ and $(0,1)$.
I tried to change the variables to $u=y-x$ and $v=1+x+y$, but then I couldn't solve the integral. Any tips on how to solve it (or the full solution) would be highly appreciated! Thanks in advance!
EDIT: I should've been more detailed in what I did. I tried the change $u=y-x$ and $v=1+x+y$, found the new domain and calculated the Jacobian. The integral then is: $$\int_1^2\int_{1-v}^{v-1}\frac{\sqrt[3]{u}}{v}\frac{1}{2}dudv$$ That is what I can't solve. I can solve it in u, but I don't even know if it's right: $$\frac{3}{8}\int_1^2\frac{{(v-1)}^{4/3}-{(1-v)}^{4/3}}{v}dv$$