Value of ∫≤ft(displaystyle1/(log x)-1/((log x)^2))dx is

The correct answer is Option (1) → \(\displaystyle\frac{x}{\log x} + c\), where c is an arbitrary constant

Let \(\log x = t\). Then \(x = e^t\) and \(dx = e^t\,dt\):

    \( \displaystyle I = \int e^t\left(\frac{1}{t} - \frac{1}{t^2}\right)dt \)

With \(f(t) = \frac{1}{t}\), we have \(f'(t) = -\frac{1}{t^2}\), so the integrand has the form \(e^t\left(f(t) + f'(t)\right)\). By the standard result \(\int e^t\left(f(t) + f'(t)\right)dt = e^t f(t) + c\):

    \( \displaystyle I = \frac{e^t}{t} + c = \frac{x}{\log x} + c \)

Shortcut: Differentiate Option (1) by the quotient rule:

    \( \displaystyle \frac{d}{dx}\left(\frac{x}{\log x}\right) = \frac{\log x - x\cdot\frac{1}{x}}{(\log x)^2} = \frac{1}{\log x} - \frac{1}{(\log x)^2} \)