Evaluate $$\int_0^1 dz \int_0^1 du \frac{u^{1+\alpha} (1-z)^{\alpha} z}{(b + z) (1-uz)},$$ for $b$ a constant. I am using the substitution $w=u(1-z)$ which gives $$\int_0^1 dz \int_0^{1-z} dw \frac{w^{\alpha}wz}{(b+z)(1-z - wz)} = \int_0^1dw \int_0^{1-w}dz \frac{w^{\alpha}wz}{(b+z)(1-z - wz)}$$ since the domain of integration is interior of triangle.
The substitution was a suggestion by a tutor in my class but I don't think it helped because I still have the coupled term $wz$. Is that the right conclusion to make?