$I = \int \dfrac{\dfrac{\sqrt{x-x^2}}{x}}{x^2} dx= \int \dfrac{\sqrt{\dfrac{1}{x}-1}}{x^2}dx$
đặt $\sqrt{\dfrac{1}{x}-1} = t \Rightarrow \dfrac{1}{x}-1=t^2 \Rightarrow \dfrac{dx}{x^2}=-2tdt$
Vậy $I = 2\int_0^{\sqrt 2} t^2 dt =\dfrac{2}{3}t^3 \bigg |_0^\sqrt 2 =\dfrac{4\sqrt 2}{3}$