Ta có $I=\int{\frac{1}{x}{{e}^{x}}dx}+\int{{{e}^{x}}\ln xdx}={{I}_{1}}+{{I}_{2}}$

Xét ${{I}_{2}}=\int{{{e}^{x}}\ln xdx}$. Đặt $\left\{ \begin{align}  & u=\ln x \\  & dv={{e}^{x}}dx \\ \end{align} \right.$$\Rightarrow \left\{ \begin{align}  & du=\frac{1}{x}dx \\  & v={{e}^{x}} \\ \end{align} \right.$

$\Rightarrow {{I}_{2}}=\left( {{e}^{x}}\ln x \right)-\int{\frac{1}{x}{{e}^{x}}dx}={{e}^{x}}\ln x-{{I}_{1}}$

Suy ra $I={{e}^{x}}\ln x+C$. Ta chọn phương án A.