Ta có

$\begin{align} & \int\limits_{0}^{1}{\left( mx+1 \right){{e}^{x}}dx}=\int\limits_{0}^{1}{\left( mx+1 \right)dx({{e}^{x}})}=\left. \left( mx+1 \right){{e}^{x}} \right|_{0}^{1}-m\int\limits_{0}^{1}{{{e}^{x}}d\left( mx+1 \right)}=\left. \left( mx+1 \right){{e}^{x}} \right|_{0}^{1}-m\int\limits_{0}^{1}{{{e}^{x}}dx} \\ & =\left[ \left( mx+1 \right){{e}^{x}} \right]_{0}^{1}-\left[ m{{e}^{x}} \right]_{0}^{1}=\left( m+1 \right)e-1-me+m=e+m-1 \\ \end{align}$

Đáp án đúng là D.