Cho hàm số $f\left( x \right)=\left\{ \begin{aligned} &...

Cách giải:
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 $t={{\ln }^{2}}x\Rightarrow dt=\dfrac{2}{x}\ln xdx\Rightarrow \dfrac{1}{x}\ln xdx=\dfrac{1}{2}dt.$
Đổi cận: $\left\{ \begin{aligned}
& x={{e}^{2}}\Rightarrow t=4 \\
& x={{e}^{4}}\Rightarrow t=16 \\
\end{aligned} \right..$
Khi đó $I=\dfrac{1}{2}\int\limits_{4}^{16}{\dfrac{f\left( t \right)}{t}dt}=\dfrac{1}{2}\int\limits_{4}^{16}{\dfrac{f\left( x \right)}{x}dx}=\dfrac{1}{2}\left( \int\limits_{4}^{8}{\dfrac{f\left( x \right)}{x}dx}+\int\limits_{8}^{16}{\dfrac{f\left( x \right)}{x}dx} \right).$
$\Rightarrow I=\dfrac{1}{2}\left( \int\limits_{4}^{8}{\dfrac{40}{x\left( x-7 \right)}dx}+\int\limits_{8}^{16}{\dfrac{{{x}^{2}}-3x}{x}dx} \right)$
$\Leftrightarrow I=\dfrac{1}{2}\left( \dfrac{40}{7}\int\limits_{4}^{8}{\left( \dfrac{1}{x-7}-\dfrac{1}{x} \right)dx}+\int\limits_{8}^{16}{\left( x-3 \right)dx} \right)$
$\Leftrightarrow I=\dfrac{20}{7}\left( \ln \left| x-7 \right|-\ln \left| x \right| \right)\left| \begin{aligned}
& 8 \\
& 4 \\
\end{aligned} \right.+\dfrac{1}{2}\left( \dfrac{{{x}^{2}}}{2}-3x \right)\left| \begin{aligned}
& 16 \\
& 8 \\
\end{aligned} \right.$
$\Leftrightarrow I=\dfrac{20}{7}\left( \ln 1-\ln 8-\ln 3+\ln 4 \right)+\dfrac{1}{2}\left( 80-8 \right)$
$\Leftrightarrow I=\dfrac{20}{7}\ln \dfrac{1}{6}+36$
$\Leftrightarrow I=36-\dfrac{20}{7}\ln 2-\dfrac{20}{7}\ln 3$