Đặt $x=e^{t}\Rightarrow dx=e^{t}dt$ và:
$x=e\Rightarrow t=1$
$x=e^{3}\Rightarrow t=3$
$\ln x=\ln e^{t}=t$
$\ln ^{2}x=\ln ^{2}e^{t}=t^{2}$
Khi đó:
$I=\int\limits_{e}^{e^{3}}\left( \dfrac{1}{\ln^{2}x}-\dfrac{1}{\ln x}\right) dx=\int\limits_{1}^{3}\left( \dfrac{1}{t^{2}}-\dfrac{1}{t}\right) e^{t}dt$
$=\int\limits_{1}^{3}\dfrac{e^{t}}{t^{2}}dt-\int\limits_{1}^{3}\dfrac{e^{t}}{t}dt$
Tính $I_{1}=\int\limits_{1}^{3}\dfrac{e^{t}}{t}dt$
Đặt $u=\dfrac{1}{t}\Rightarrow du=-\dfrac{dt}{t^{2}}$
$dv=e^{t}dt\Rightarrow v=e^{t}$
Khi đó $I_{1}=\dfrac{e^{t}}{t}\left|\begin{matrix} 3 \\ \\ 1\end{matrix}\right. +\int\limits_{1}^{3}\dfrac{e^{t}}{t^{2}}dt$
$I_{1}=\dfrac{e^{3}}{3}-e+\int\limits_{1}^{3}\dfrac{e^{t}}{t^{2}}dt$
Khi đó $I=\int\limits_{1}^{3}\dfrac{e^{t}}{t^{2}}dt-I_{1}$
$=\int\limits_{1}^{3}\dfrac{e^{t}}{t^{2}}dt-\left( \dfrac{e^{3}}{3}-e+\int\limits_{1}^{3}\dfrac{e^{t}}{t^{2}}dt\right)$
$=e-\dfrac{e^{3}}{3}=\dfrac{3e-e^{3}}{3}$