$\int\frac{\log_e x}{(1+ \log_e x)^2}dx$ is equal to

$\frac{x}{1+ \log_e x}+C$, where C is constant of integration

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

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

Note: $\frac{d}{dx}\!\left(\frac{x}{1+\log_e x}\right)=\frac{(1+\log_e x)-1}{(1+\log_e x)^2}=\frac{\log_e x}{(1+\log_e x)^2}$

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

$\frac{x}{1+\log_e x}+C$

DIRECT SOLUTION

$\int \frac{\ln x}{(1+\ln x)^2} \, dx$

$t = 1 + \ln x \Rightarrow dt = \frac{1}{x} dx$

$dx = x \, dt$

$\ln x = t - 1,\quad x = e^{t-1}$

$\int \frac{t-1}{t^2} \cdot e^{t-1} dt$

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

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

$= e^{-1} \int \frac{d}{dt}\left(\frac{e^t}{t}\right) dt$

$= e^{-1} \cdot \frac{e^t}{t} + C$

$= \frac{e^{t-1}}{t} + C$

$= \frac{x}{1+\ln x} + C$

The value is $\frac{x}{1+\ln x} + C$.