Tính 

D

- Hướng dẫn giải

Chọn D

Đặt

$\left\{\begin{array}{l}u=x\\ dv=e^{2x}dx\end{array}\Rightarrow \left\{\begin{array}{l}du=dx\\ v=\frac{1}{2}{e}^{2x}\end{array}\right.\right.$

Khi đó 

$$\begin{gathered} {\int }_0^{1}x.e^{2x}dx=\frac{1}{2}x.e^{2x}{|}_0^1-\frac{1}{2}{\int }_0^1{e}^{2x}dx \\ =\frac{1}{2}x.e^{2x}{|}_0^1-\frac{1}{4}{e}^{2x}{|}_0^1 \\ =\frac{1}{2}{e}^2-\frac{1}{4}{e}^2+\frac{1}{4}=\frac{1+e^2}{4} \end{gathered}$$