Question

$\underset{e}{\overset{e^2}{\int }}\left\{\frac{1}{\log x}-\frac{1}{{\left(\log x\right)}^2}\right\}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_e^{e^2}\left\{\frac{1}{\log x}-\frac{1}{{\left(\log x\right)}^2}\right\}\mathit{d}x.\mathrm{Then}, \\ I={\int }_e^{e^2}1\frac{1}{\log x}dx-{\int }_e^{e^2}\frac{1}{{\left(\log x\right)}^2}dx \\ \mathrm{Integrating}\mathrm{by}\mathrm{parts} \\ \Rightarrow I=\left\{{\left[\frac{x}{\log x}\right]}_e^{e^2}-{\int }_e^{e^2}\frac{-1}{x{\left(\log x\right)}^2}x\mathit{d}x\right\}-{\int }_e^{e^2}\frac{1}{{\left(\log x\right)}^2}dx \\ \Rightarrow I={\left[\frac{x}{\log x}\right]}_e^{e^2}+{\int }_e^{e^2}\frac{1}{{\left(\log x\right)}^2}\mathit{d}x-{\int }_e^{e^2}\frac{1}{{\left(\log x\right)}^2}dx \\ \Rightarrow I={\left[\frac{x}{\log x}\right]}_e^{e^2}+0 \\ \Rightarrow I=\frac{e^2}{\log e^2}-\frac{e}{\log e} \\ \Rightarrow I=\frac{e^2}{2\log e}-\frac{e}{\log e} \\ \Rightarrow I=\frac{e^2}{2}-e \end{gathered}$$