$I=\int_0^4 \dfrac{x^2}{1+x\sqrt x}dx =\dfrac{1}{3}\int_0^4 \dfrac{3x^2}{1+\sqrt{x^3}}dx$
đặt $\sqrt{x^3} =t \Rightarrow x^3 = t^2 \Rightarrow 3x^2 dx = 2tdt$
Vậy $I=2\int_0^8 \dfrac{t}{1+t}dt=2\int \bigg ( 1 -\dfrac{1}{t+1} \bigg ) dt =2(t -\ln|t+1| ) \bigg |_0^8=16-2\ln 9$