$\int_0^1 \log (x+1) \mathrm{d} x=$

The definite integral $\int_0^1 \log (x+1) \, dx$ is evaluated using integration by parts.

Choosing $u = \log(x+1)$ and $dv = dx$ gives $du = \frac{1}{x+1} dx$ and $v = x$. Applying the formula yields:

$\int_0^1 \log (x+1) \, dx = [x \log(x+1)]_0^1 - \int_0^1 \frac{x}{x+1} \, dx$.

The boundary term simplifies entirely to $\log 2$. The remaining integral is simplified by writing the integrand as $1 - \frac{1}{x+1}$, leading to:

$\int_0^1 \left(1 - \frac{1}{x+1}\right) dx = [x - \log(x+1)]_0^1 = 1 - \log 2$.

Combining both results:

$\int_0^1 \log (x+1) \, dx = \log 2 - (1 - \log 2) = 2\log 2 - 1$.

This corresponds to option D.