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B corresponds to the triangle with vertices (0,0), (1,0) and (0,1)

I believe I am supposed to use variable substitution; however, whenever I can find a way to simplify the function, the integration area becomes horrendous, so what would be the best approach for this integration? I tried the most obvious substitution, $u=y-x$ and $v=1+y+x$, what resulted is a slightly more complicated triangle, and a similarly difficult function, more specifically $$ \int_{1}^{2}\int_{1-v}^{1+v}\frac{\sqrt[3]{u}}{v}\,du\,dv\rightarrow \int_{1}^{2}\frac{1+v}{v}\sqrt[3]{1+v}-\frac{1-v}{v}\sqrt[3]{1-v}\, dv$$ If I did everything correctly, of course.

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  • $\begingroup$ The integral is trivially zero because the given integration domain is invariant with respect to $(x,y)\mapsto(y,x)$, but about the integrand function $f(y,x)=\color{red}{-} f(x,y)$. $\endgroup$ – Jack D'Aurizio Apr 21 '18 at 14:05
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Hint:

You may break down the region of integration $B$ as $B_x\cup B_y\cup B_d$ where $B_x:=\{(x,0)\mid x\in [0,1]\}$, $B_y:=\{(0,y)\mid y\in [0,1]\}$ and $B_d:=\{(k,1-k)\mid k\in [0,1]\}$

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