$I = \int \dfrac{\dfrac{\sqrt{x-x^2}}{x}}{x^2} dx= \int \dfrac{\sqrt{\dfrac{1}{x}-1}}{x^2}dx$đặt $\sqrt{\dfrac{1}{x}-1} = t \Rightarrow \dfrac{1}{x}-1=t^2 \Rightarrow \dfrac{dx}{x^2}=-2tdt$Vậy $I = 2\int_0^{\sqrt 2} t^2 dt =\dfrac{2}{3}t^3 \bigg |_0^\sqrt 2 =\dfrac{4\sqrt 2}{3}$
$I = \int \dfrac{\dfrac{\sqrt{x-x^2}}{x}}{x^2} dx= \int \dfrac{\sqrt{\dfrac{1}{x}-1}}{x^2}dx$đặt $\sqrt{\dfrac{1}{x}-1} = t \Rightarrow \dfrac{1}{x}-1=t^2 \Rightarrow \dfrac{dx}{x^2}=-2tdt$Vậy $I = 2\int_{\frac{1}{3}}^1 t^2 dt =\dfrac{2}{3}t^3 \bigg |_{\frac{1}{3}}^1 =\dfrac{4\sqrt 2}{3}$
$I = \int \dfrac{\dfrac{\sqrt{x-x^2}}{x}}{x^2} dx= \int \dfrac{\sqrt{\dfrac{1}{x}-1}}{x^2}dx$đặt $\sqrt{\dfrac{1}{x}-1} = t \Rightarrow \dfrac{1}{x}-1=t^2 \Rightarrow \dfrac{dx}{x^2}=-2tdt$Vậy $I = 2\int_
0^{\
sqr
t 2} t^2 dt =\dfrac{2}{3}t^3 \bigg |_
0^\
sqr
t 2 =\dfrac{4\sqrt 2}{3}$