Question

Evaluate the following definite integrals $(1-58)$:

${\int }_e^{e^2}\{\frac{1}{\log x}-\frac{1}{(\log x{)}^2}\}dx$

Answer 1

$I={\int }_e^{e^2}\frac{1}{\log x}.1{\int }_e^{e^2}\frac{1}{(\log x{)}^2}dx={\left[\frac{x}{\log x}\right]}_e^{e^2}-{\int }_e^{e^2}\frac{1}{x(\log x{)}^2}xdx-{\int }_e^{e^2}\frac{1}{(\log x{)}^2}dx$

$\Rightarrow I=\frac{e^2}{\log e^2}-\frac{e}{\log e}=\frac{e^2}{2\log e}=\frac{e}{\log e}=\frac{e^2}{2}-e$