Gợi ý $I=\int \dfrac{\sqrt[3]{\dfrac{x-x^3}{x^3}}}{x^3}dx=-\dfrac{1}{2}\int \dfrac{-2. \sqrt[3]{\dfrac{1}{x^2}-1}}{x^3}dx$
Đặt $\sqrt[3]{\dfrac{1}{x^2}-1}=t\Rightarrow \dfrac{1}{x^2}-1=t^3 \Rightarrow -\dfrac{2}{x^3}dx=3t^2dt$
Vậy $I=-\dfrac{1}{2}\int 3t^2 .tdt =-\frac{3}{2}\int t^3 dt$ dễ rồi tự làm nhé