I found this integral in a FB page: $$I=\int_{0}^{1}\int_{0}^{1}\frac{\log{(1+x^3+y^3)}-\log{(xy)}}{1+x^3+y^3}dxdy$$
I am trying to evaluate it, but it is hard.
My attemps: convert by using substitution $x=r\cos^{2/3}({\theta}), y=r\sin^{2/3}({\theta})$ then it becomes: $$I=\frac{4}{3}\int_{0}^{\frac{\pi}{4}}\int_{0}^{\frac{1}{\cos^{2/3}({\theta})}}\frac{\log{(1+r^3)}-\log{(r^2\cos^{2/3}({\theta})\sin^{2/3}({\theta}))}}{(1+r^3)\cos^{1/3}({\theta})\sin^{1/3}({\theta})}dxdy$$
And it seems impossible due to the antiderivative is a mess.
I also try to transform: $x^3+y^3=(x+y)^3-3xy(x+y)$ so i set $u=x+y, v=xy$ but it is still hard.
Another try, because of symmetry then the integral can be rewritten as: $$I=\int_{0}^{1}\int_{0}^{1}\frac{\log{(1+x^3+y^3)}-\log{(x^2)}}{1+x^3+y^3}dxdy$$ But it doesn't help much. I also add more variable like this: $$I=\int_{0}^{1}\int_{0}^{1}\int_{0}^{\infty}(\log{(1+x^3+y^3)}-\log{(x^2)})e^{-z(1+x^3+y^3)}dxdydz$$ And it leads to something messier. So, I need some helps from every one, thank you for your time.
