$\int_0^1 \frac{1}{2+\sqrt{x}} \mathrm{~d} x=$

The integral $\int_0^1 \frac{1}{2+\sqrt{x}} dx$ is evaluated using the substitution $u = \sqrt{x}$, which gives $u^2 = x$ and $dx = 2u\,du$.

With the limits transforming as $u(0) = 0$ and $u(1) = 1$, the integral becomes $\int_0^1 \frac{2u}{2+u} du$.

Rewriting the integrand algebraically, $\frac{2u}{2+u} = 2 - \frac{4}{u+2}$.

Integrating term by term yields $\left[2u - 4\ln|u+2|\right]_0^1 = (2 - 4\ln 3) - (0 - 4\ln 2) = 2 + 4(\ln 2 - \ln 3) = 2 + 4\ln\left(\frac{2}{3}\right)$.

Option B evaluates to $2\ln\left(\frac{4e}{9}\right) = 2(\ln 4 + 1 - \ln 9) = 2(2\ln 2 + 1 - 2\ln 3) = 4\ln 2 + 2 - 4\ln 3 = 2 + 4\ln\left(\frac{2}{3}\right)$, confirming the match.

Final answer: $\boxed{\text{B}}$