Đặt \(\left\{ \begin{array}{l}u = \ln x\\dv = \dfrac{1}{{{x^2}}}dx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = \dfrac{1}{x}dx\\v = - \dfrac{1}{x}\end{array} \right.\)
\( \Rightarrow \int\limits_1^e {\dfrac{{\ln x}}{{{x^2}}}dx} = \left. { - \dfrac{{\ln x}}{x}} \right|_1^e + \int\limits_1^e {\dfrac{1}{{{x^2}}}dx} \) \( = - \dfrac{1}{e} - \left. {\dfrac{1}{x}} \right|_1^e = - \dfrac{1}{e} - \dfrac{1}{e} + 1\) \( = - \dfrac{2}{e} + 1 = 1 - \dfrac{2}{e}\).
Chọn B.
