\( \int\displaystyle\left\{\frac{1}{\log_e x}-\frac{1}{(\log_e x)^2}\right\}dx \) is equal to : (where \( C \) is an arbitrary constant)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{x}{\log_e x}+C \)
Integrating the first part by parts, with \( \frac{1}{\log_e x} \) as the first function and 1 as the second:
\( \displaystyle \int\frac{1}{\log_e x}\,dx=\frac{x}{\log_e x}-\int x\cdot\left(-\frac{1}{(\log_e x)^2}\cdot\frac{1}{x}\right)dx \)
\( \displaystyle =\frac{x}{\log_e x}+\int\frac{dx}{(\log_e x)^2} \)
The last integral cancels the second part of the given integral, so
\( \displaystyle \int\left\{\frac{1}{\log_e x}-\frac{1}{(\log_e x)^2}\right\}dx=\frac{x}{\log_e x}+C \)
Shortcut: Differentiate Option (2):
\( \displaystyle \frac{d}{dx}\left(\frac{x}{\log_e x}\right) \)
\( \displaystyle =\frac{\log_e x-x\cdot\frac{1}{x}}{(\log_e x)^2} \)
\( \displaystyle =\frac{1}{\log_e x}-\frac{1}{(\log_e x)^2} \)