∫≤ft(1/(log_e t)-1/((log_e t)^2)) dt is equal to

The correct answer is Option (2) → \(\displaystyle\frac{t}{\log_e t}+C\), where C is constant of integration

Let \(t = e^u\), so \(\log_e t = u\) and \(dt = e^u\,du\):

    \( \displaystyle \int\left(\frac{1}{u} - \frac{1}{u^2}\right) e^u\,du \)

This has the form \(\int e^u\left[f(u) + f'(u)\right] du = e^u f(u) + C\) with \(f(u) = \frac{1}{u}\), since \(f'(u) = -\frac{1}{u^2}\):

    \( \displaystyle = \frac{e^u}{u} + C \)

    \( \displaystyle = \frac{t}{\log_e t} + C \)

Shortcut: Differentiate Option (2) by the quotient rule:

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

    \( \displaystyle = \frac{1}{\log_e t} - \frac{1}{(\log_e t)^2} \)