∫≤ft(displaystyle1/(log_e x)-1/((log_e x)^2))dx is equal to

The correct answer is Option (1) → \(\displaystyle\frac{x}{\log_e x}+C\): C is an arbitrary constant

The first part is integrated by integration by parts, taking \(\frac{1}{\log_e x}\) as the first function and 1 as the second:

    \(\displaystyle \int \frac{1}{\log_e x}\,dx = \frac{1}{\log_e x}\cdot x - \int \left(-\frac{1}{(\log_e x)^2}\cdot\frac{1}{x}\right) x\,dx\)

    \(\displaystyle = \frac{x}{\log_e x} + \int \frac{1}{(\log_e x)^2}\,dx\)

Putting this into the given integral, the two integrals of \(\frac{1}{(\log_e x)^2}\) cancel:

    \(\displaystyle \int\left(\frac{1}{\log_e x}-\frac{1}{(\log_e x)^2}\right)dx = \frac{x}{\log_e x} + C\)

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

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

This is the integrand, so Option (1) is the integral.