The value of \(\int\displaystyle\left\{\frac{1}{\log_e x}-\frac{1}{(\log_e x)^2}\right\} dx\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{x}{\log_e x} + c\): where c is an arbitrary constant
By integration by parts on the first term, with \(\frac{1}{\log x}\) as the first function and 1 as the second:
\(\displaystyle \int \frac{1}{\log x}\,dx = \frac{1}{\log x}\cdot x - \int x\cdot\left(-\frac{1}{(\log x)^2}\cdot\frac{1}{x}\right)dx\)
\(\displaystyle = \frac{x}{\log x} + \int \frac{1}{(\log x)^2}\,dx\)
The last integral cancels the second term of the integrand:
\(\displaystyle \int\left\{\frac{1}{\log x} - \frac{1}{(\log x)^2}\right\}dx = \frac{x}{\log x} + c\)
Shortcut: differentiate Option (4) by the quotient rule:
\(\displaystyle \frac{d}{dx}\left(\frac{x}{\log x}\right) = \frac{\log x - x\cdot\frac{1}{x}}{(\log x)^2} = \frac{1}{\log x} - \frac{1}{(\log x)^2}\)