Phương pháp giải:

\(F\left( x \right) = \int\limits_{}^{} {f\left( x \right)dx} \).

Lời giải chi tiết:

\(\begin{array}{l}F\left( x \right) = \int\limits_{}^{} {f\left( x \right)dx}  = \int\limits_{}^{} {\dfrac{{dx}}{{x\ln x}}}  = \int\limits_{}^{} {\dfrac{{d\left( {\ln x} \right)}}{{\ln x}}}  = \ln \left| {\ln x} \right| + C\\F\left( x \right) = \left[ \begin{array}{l}\ln \left( {\ln x} \right) + {C_1}\,\,khi\,\,x \ge 1\\\ln \left( { - \ln x} \right) + {C_2}\,\,khi\,\,0 < x < 1\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}F\left( {\dfrac{1}{e}} \right) = 2\\F\left( e \right) = \ln 2\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{C_2} = 2\\{C_1} = \ln 2\end{array} \right.\\ \Rightarrow F\left( x \right) = \left[ \begin{array}{l}\ln \left( {\ln x} \right) + \ln 2\,\,khi\,\,x \ge 1\\\ln \left( { - \ln x} \right) + 2\,\,khi\,\,0 < x < 1\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}F\left( {\dfrac{1}{{{e^2}}}} \right) = \ln \left( { - \ln \dfrac{1}{{{e^2}}}} \right) + 2 = \ln 2 + 2\\F\left( {{e^2}} \right) = \ln \left( {\ln {e^2}} \right) + \ln 2 = 2\ln 2\end{array} \right.\\ \Rightarrow F\left( {\dfrac{1}{{{e^2}}}} \right) + F\left( {{e^2}} \right) = 3\ln 2 + 2\end{array}\)

Chọn D.