Ta có $\dfrac{x^2 +x}{\sqrt[3]{(x+1)^2}}=\dfrac{x(x+1)}{(x+1)^{\frac{2}{3}}}=x(x+1)^{\frac{1}{3}}=x \sqrt[3]{x+1}$
$I=\int x \sqrt[3]{x+1}dx$ đặt $\sqrt[3]{x+1}=t \Rightarrow dx = 3t^2 dt$
$I=\int_1^{\sqrt[3]{2}} (t^3-1) t .3t^2 dt =3\int (t^6 -t^3)dt$ dễ tự làm nốt