$\int \frac{1}{e^x+1} \mathrm{~d} x=$

The integral $\int \frac{1}{e^x+1} \, dx$ can be evaluated by rewriting the integrand algebraically:

$\frac{1}{e^x+1} = \frac{e^x+1-e^x}{e^x+1} = 1 - \frac{e^x}{e^x+1}$

This transformation simplifies the integral to:

$\int \left(1 - \frac{e^x}{e^x+1}\right) \, dx = \int 1 \, dx - \int \frac{e^x}{e^x+1} \, dx$

The first term integrates directly to $x$, while the second requires substitution. Let $u = e^x+1$, so $du = e^x \, dx$

Substituting yields:

$\int \frac{e^x}{e^x+1} \, dx = \int \frac{1}{u} \, du = \log u = \log(e^x+1)$

The complete integral is therefore:

$x - \log(e^x+1) + C$

This result corresponds to option B.

Answer: $\boxed{\text{B}}$