Question

\[\int\limits_e^{e^2} \left\{ \frac{1}{\log x} - \frac{1}{\left( \log x \right)^2} \right\} dx\]

Answer 1

\[Let\ I = \int_e^{e^2} \left\{ \frac{1}{\log x} - \frac{1}{\left( \log x \right)^2} \right\} d x . Then, \]
\[I = \int_e^{e^2} 1 \frac{1}{\log x} dx - \int_e^{e^2} \frac{1}{\left( \log x \right)^2} dx\]
\[\text{Integrating by parts}\]
\[ \Rightarrow I = \left\{ \left[ \frac{x}{\log x} \right]_e^{e^2} - \int_e^{e^2} \frac{- 1}{x \left( \log x \right)^2} x d x \right\} - \int_e^{e^2} \frac{1}{\left( \log x \right)^2} dx\]
\[ \Rightarrow I = \left[ \frac{x}{\log x} \right]_e^{e^2} + \int_e^{e^2} \frac{1}{\left( \log x \right)^2} d x - \int_e^{e^2} \frac{1}{\left( \log x \right)^2} dx\]
\[ \Rightarrow I = \left[ \frac{x}{\log x} \right]_e^{e^2} + 0\]
\[ \Rightarrow I = \frac{e^2}{\log e^2} - \frac{e}{\log e}\]
\[ \Rightarrow I = \frac{e^2}{2 \log e} - \frac{e}{\log e}\]
\[ \Rightarrow I = \frac{e^2}{2} - e\]