Đáp án đúng: A Đặt $\left\{ \begin{array}{l}u={{x}^{2}}+1\\dv={{e}^{x}}dx\end{array} \right.\Rightarrow \left\{ \begin{array}{l}du=2xdx\\v={{e}^{x}}\end{array} \right.$ nên $\begin{array}{l}{{I}_{14}}=\left. \left( {{x}^{2}}+1 \right){{e}^{x}} \right|_{0}^{1}-\int\limits_{0}^{1}{2{{e}^{x}}xdx}\\=\left. \left( {{x}^{2}}+1 \right){{e}^{x}} \right|_{0}^{1}-2=2e-3.\end{array}$ Vì xét$\int\limits_{0}^{1}{x{{e}^{x}}dx}=\left. x{{e}^{x}} \right|_{0}^{1}-\int\limits_{0}^{1}{{{e}^{x}}dx}=1.$