Evaluate the integral

$\begin{aligned} I & = \int \frac 1{e^x + 1} dx \\ & = \int \frac {e^x+1 - e^x}{e^x+1} dx \\ & = \int \left(1 - \frac {e^x}{e^x+1} \right) dx \\ & = \boxed {x - \ln(e^x +1) + \blue C} & \small \blue{\text{where }C \text{ is the constant of integration.}} \end{aligned}$

Alternatively:

Since the answer options are given, we can also check which answer is correct as follows:

$\begin{cases} \dfrac d{dx} \left(x - \ln (e^x + 1) + C \right) = 1 - \dfrac {e^x}{e^x+1} = \dfrac 1{e^x+1} & \small \blue{\text{The correct answer}} \\ \dfrac d{dx} \left(\dfrac {(e^x + 1)^2}2 + C \right) = e^x(e^x+1) & \small \red{\text{Not the answer}} \end{cases}$

The value of the integral is $\ln \left (1-\dfrac {1}{e^x+1}\right ) +C=x-\ln (1+e^x)+C$ , where $C$ is the integration constant.