Chọn: Đáp án D

$I=\int\limits_{1}^{e}{\frac{1-\ln x}{x}dx}=\int\limits_{1}^{e}{\frac{1}{x}dx}-\int\limits_{1}^{e}{\frac{\ln x}{x}dx}={{I}_{1}}-{{I}_{2}}$

${{I}_{1}}=\int\limits_{1}^{e}{\frac{1}{x}dx}=\ln \left| x \right|\left| \begin{align}  & e \\  & 1 \\ \end{align} \right.=1;{{I}_{2}}=\int\limits_{1}^{e}{\frac{\ln x}{x}dx}$

Đặt $t=\ln x\Rightarrow dt=\frac{1}{x}dx$. Đổi cận$x=1\Rightarrow t=0;x=e\Rightarrow t=1\Rightarrow {{I}_{2}}=\int\limits_{0}^{1}{tdt}=\frac{{{t}^{2}}}{2}\left| \begin{align} & 1 \\  & 0 \\ \end{align} \right.=\frac{1}{2}$

Vậy $I={{I}_{1}}-{{I}_{2}}=\frac{1}{2}$