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Question

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

Solution

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

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Definite Integrals
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Chapter 20: Definite Integrals - Exercise 20.1 [Page 17]

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RD Sharma Mathematics [English] Class 12
Chapter 20 Definite Integrals
Exercise 20.1 | Q 36 | Page 17

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