∫(log_e x)/((1+ log_e x)^2)dx is equal to

The correct answer is Option (2) → \(\displaystyle\frac{x}{1+ \log_e x}+C\), where C is constant of integration

Let \(t = \log_e x\), so \(x = e^t\) and \(dx = e^t\,dt\):

    \(\displaystyle I = \int e^t\,\frac{t}{(1 + t)^2}\,dt\)

Writing the numerator as \(t = (1 + t) - 1\):

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

This has the form \(\int e^t[f(t) + f'(t)]\,dt = e^t f(t) + C\), with

    \(\displaystyle f(t) = \frac{1}{1 + t},\quad f'(t) = -\frac{1}{(1 + t)^2}\)

So

    \(\displaystyle I = \frac{e^t}{1 + t} + C = \frac{x}{1 + \log_e x} + C\)

Shortcut: Differentiate the option. By the quotient rule,

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

which is the integrand.