\(\int\left\{\displaystyle\frac{1}{\log x}-\frac{1}{(\log x)^2}\right\}dx\), where \(x \gt 1\), is equal to (C is a constant of integration)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{x}{\log x}+C\)
Integration by parts is applied to the first part only, with \(\frac{1}{\log x}\) as the first function and 1 as the second:
\( \displaystyle \frac{d}{dx}\left(\frac{1}{\log x}\right) = -\frac{1}{x(\log x)^2} \)
\( \displaystyle \int\frac{dx}{\log x} = \frac{x}{\log x}+\int x\cdot\frac{1}{x(\log x)^2}\,dx \)
\( \displaystyle = \frac{x}{\log x}+\int\frac{dx}{(\log x)^2} \)
So the second part of the integrand cancels:
\( \displaystyle \int\left\{\frac{1}{\log x}-\frac{1}{(\log x)^2}\right\}dx = \frac{x}{\log x}+C \)
Check: by the quotient rule,
\( \displaystyle \frac{d}{dx}\left(\frac{x}{\log x}\right) = \frac{\log x-1}{(\log x)^2} = \frac{1}{\log x}-\frac{1}{(\log x)^2} \)