Letting $t=\frac{x+1}{x}\iff \frac 1x=t-1$, then we get $$\frac{dt}{dx}=\frac{x-(x+1)}{x^2}=-\frac{1}{x^2}=-(t-1)^2.$$ Hence, we have $$\int\frac{1}{x+1}\left(\frac{x+1}{x}\right)^{2/3}dx=\int\frac{1}{\frac{1}{t-1}+1}\cdot t^{2/3}\cdot \frac{-1}{(t-1)^2}dt=\int\frac{-t^{2/3}}{(t-1)+(t-1)^2}dt$$ $$=\int\frac{-t^{2/3}}{(t-1)t}dt=\int\frac{-t^{-(1/3)}}{t-1}dt.$$
Then, this gives you the answer.
So, the answer is $$\frac 12\ln (x^{2/3}+x^{1/3}+1)-\ln (1-x^{1/3})-\sqrt 3\tan^{-1}\left(\frac{2x^{1/3}+1}{\sqrt 3}\right)+C.$$