integral $\int \frac{1}{x\left(2+\sqrt[3]{\frac{x-1}{x}}\right)} dx$

Question

$$\int \frac{1}{x\left(2+\sqrt[3]{\frac{x-1}{x}}\right)} dx$$ My answer is $$ \textrm{let} \,u = \sqrt[3]{\frac{x-1}{x}} \Rightarrow \frac{x-1}{x} = u^{3} $$ $$ \Rightarrow x = \frac{1}{1-u^{3}} \Rightarrow dx = \frac{3u^{2}}{(1-u^{3})^{2}}du $$ $$ \Rightarrow I = \int \frac{(1-u^{2})(3u^{2})}{(2+u)(1-u^{3})^{2}}du \Rightarrow I = 3\int \frac{u^{2}}{(2+u)(1-u^{3})}$$ I know that it can be solved by partial fraction by i don't like it ... Is $u$-substitution not completely right or is there any easier method to solve it ..

Answer 1

HINT: convert your Integrand into $$-1/3\, \left( u-1 \right) ^{-1}+{\frac {-u-1}{{u}^{2}+u+1}}+4/3\, \left( 2+u \right) ^{-1} $$