Đáp án đúng: C
Giải chi tiết:Ta có \(I = \int\limits_{{e^2}}^{{e^4}} {\dfrac{{f\left( {{{\ln }^2}x} \right)}}{{x\ln x}}dx}  = \int\limits_{{e^2}}^{{e^4}} {\dfrac{{f\left( {{{\ln }^2}x} \right)\ln x}}{{x{{\ln }^2}x}}dx} \)Đặt \(t = {\ln ^2}x\,\, \Rightarrow dt = \dfrac{2}{x}\ln x.dx\, \Rightarrow \dfrac{1}{x}\ln xdx = \dfrac{1}{2}dt\).Đổi cận: \(\left\{ \begin{array}{l}x = {e^2} \Rightarrow t = 4\\x = {e^4} \Rightarrow t = 16\end{array} \right.\).Khi đó \(I = \dfrac{1}{2}\int\limits_4^{16} {\dfrac{{f\left( t \right)}}{t}dt}  = \dfrac{1}{2}\int\limits_4^{16} {\dfrac{{f\left( x \right)}}{x}dx}  = \dfrac{1}{2}\left( {\int\limits_4^8 {\dfrac{{f\left( x \right)}}{x}dx}  + \int\limits_8^{16} {\dfrac{{f\left( x \right)}}{x}dx} } \right)\).\(\begin{array}{l} \Rightarrow I = \dfrac{1}{2}\left( {\int\limits_4^8 {\dfrac{{40}}{{x\left( {x - 7} \right)}}dx}  + \int\limits_8^{16} {\dfrac{{{x^2} - 3x}}{x}dx} } \right)\\ \Leftrightarrow I = \dfrac{1}{2}\left( {\dfrac{{40}}{7}\int\limits_4^8 {\left( {\dfrac{1}{{x - 7}} - \dfrac{1}{x}} \right)dx}  + \int\limits_8^{16} {\left( {x - 3} \right)dx} } \right)\\ \Leftrightarrow I = \dfrac{{20}}{7}\left. {\left( {\ln \left| {x - 7} \right| - \ln \left| x \right|} \right)} \right|_4^8 + \left. {\dfrac{1}{2}\left( {\dfrac{{{x^2}}}{2} - 3x} \right)} \right|_8^{16}\\ \Leftrightarrow I = \dfrac{{20}}{7}\left( {\ln 1 - \ln 8 - \ln 3 + \ln 4} \right) + \dfrac{1}{2}\left( {80 - 8} \right)\\ \Leftrightarrow I = \dfrac{{20}}{7}\ln \dfrac{1}{6} + 36\\ \Leftrightarrow I = 36 - \dfrac{{20}}{7}\ln 2 - \dfrac{{20}}{7}\ln 3\end{array}\)Chọn C