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

Xét $I=\int\limits_{e}^{{{e}^{2}}}{\dfrac{f({{\ln }^{2}}x)}{x\ln x}}dx$.
Đặt $u={{\ln }^{2}}x$ $\Rightarrow du=\dfrac{2\ln x}{x}dx=\dfrac{2{{\ln }^{2}}x}{x\ln x}dx=\dfrac{2u}{x\ln x}dx\Rightarrow \dfrac{dx}{x\ln x}=\dfrac{du}{2u}.$
Đổi cận : $\left\{ \begin{aligned}
& x=e\Rightarrow u=1 \\
& x={{e}^{2}}\Rightarrow u=4 \\
\end{aligned} \right.$.
Khi đó
$\begin{aligned}
& I=\dfrac{1}{2}\int\limits_{1}^{4}{\dfrac{f(u)}{u}}du=\dfrac{1}{2}\int\limits_{1}^{4}{\dfrac{f(x)}{x}}dx=\dfrac{1}{2}\left( \int\limits_{1}^{2}{\dfrac{f(x)}{x}}dx+\int\limits_{2}^{4}{\dfrac{f(x)}{x}}dx \right) \\
& =\dfrac{1}{2}\left( \int\limits_{1}^{2}{\dfrac{2}{x\left( 2x-5 \right)}}dx+\int\limits_{2}^{4}{\dfrac{3{{x}^{2}}+6x}{x}}dx \right)=\dfrac{1}{2}\left( \int\limits_{1}^{2}{\dfrac{2}{x\left( 2x-5 \right)}}dx+\int\limits_{2}^{4}{\left( 3x+6 \right)}dx \right) \\
& =\dfrac{1}{2}\left[ \dfrac{4}{5}\int\limits_{1}^{2}{\left( \dfrac{1}{2x-5}-\dfrac{1}{2x} \right)}dx+\left. \left( \dfrac{3{{x}^{2}}}{2}+6x \right) \right|_{2}^{4} \right]=\dfrac{1}{2}\left[ \dfrac{4}{5}.\dfrac{1}{2}\left. \ln \left| \dfrac{2x-5}{2x} \right| \right|_{1}^{2}+30 \right] \\
& =\dfrac{1}{2}\left[ \dfrac{2}{5}\left( -\ln 6 \right)+30 \right]=15-\dfrac{1}{5}\ln 6 \\
\end{aligned}$.