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

Evaluate the integral: $\int \frac{\mathrm{d}x}{e^x - 1}$

Let $u = e^x - 1$, so that $\mathrm{d}u = e^x \mathrm{d}x$ and $e^x = u + 1$. The integral becomes:

$\int \frac{1}{u} \cdot \frac{\mathrm{d}u}{u + 1} = \int \frac{1}{u(u+1)} \mathrm{d}u$

Decompose into partial fractions: $\frac{1}{u(u+1)} = \frac{1}{u} - \frac{1}{u+1}$

Integrating:

$\int \left( \frac{1}{u} - \frac{1}{u+1} \right) \mathrm{d}u = \ln|u| - \ln|u+1| + C = \ln\left| \frac{u}{u+1} \right| + C$

Substituting $u = e^x - 1$:

$\ln\left| \frac{e^x - 1}{e^x} \right| + C = \ln|e^x - 1| - \ln|e^x| + C$

For real $x$, $e^x > 0$ so $\ln|e^x| = x$, giving:

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

Result: This corresponds to option B.