Câu hỏi 24

Ta có: \(I = \int {x\sqrt {1 - 2x} dx} .\)

Đặt \(t = \sqrt {1 - 2x}  \Rightarrow {t^2} = 1 - 2x \Rightarrow 2tdt =  - 2dx \Rightarrow dx =  - tdt.\)

\(\begin{array}{l} \Rightarrow x = \dfrac{{1 - {t^2}}}{2}.\\ \Rightarrow I =  - \int {\dfrac{{1 - {t^2}}}{2}.{t^2}dt}  = \dfrac{1}{2}\int {\left( {{t^4} - {t^2}} \right)dt}  = \dfrac{1}{2}\left( {\dfrac{{{t^5}}}{5} - \dfrac{{{t^3}}}{3}} \right) + C\\ = \dfrac{{{t^5}}}{{10}} - \dfrac{{{t^3}}}{6} + C = \dfrac{{\sqrt {{{\left( {1 - 2x} \right)}^5}} }}{{10}} - \dfrac{{\sqrt {{{\left( {1 - 2x} \right)}^3}} }}{6} + C.\end{array}\)

Chọn C.