Câu hỏi 23

Ta có: \(I = \int {\left( {1 - x} \right){e^{2x}}dx} \) 

Đặt \(\left\{ \begin{array}{l}u = 1 - x\\dv = {e^{2x}}dx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du =  - dx\\v = \dfrac{1}{2}{e^{2x}}\end{array} \right.\)

\( \Rightarrow I = \dfrac{{\left( {1 - x} \right){e^{2x}}}}{2} + \int {\dfrac{1}{2}{e^{2x}}dx} \) \( = \dfrac{{\left( {1 - x} \right){e^{2x}}}}{2} + \dfrac{1}{4}{e^{2x}} + C = \dfrac{{\left( {3 - 2x} \right){e^{2x}}}}{4} + C.\)

Chọn  B.