$\int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} \frac{x}{1+\sin x} \mathrm{~d} x=$

Given the integral $I = \int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} \frac{x}{1+\sin x} \, dx$, apply the transformation property $\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx$ with $a+b = \pi$:

$I = \int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} \frac{\pi-x}{1+\sin x} \, dx$

Adding this to the original expression yields:

$2I = \int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} \frac{\pi}{1+\sin x} \, dx$

Rewriting the integrand through rationalization:

$\frac{1}{1+\sin x} = \frac{1-\sin x}{\cos^2 x} = \sec^2 x - \sec x \tan x$

Integration gives:

$\int \left(\sec^2 x - \sec x \tan x\right) \, dx = \tan x - \sec x + C$

Evaluating the definite integral:

$2I = \pi \left[ \tan x - \sec x \right]_{\frac{\pi}{3}}^{\frac{2\pi}{3}}$

$\tan\left(\frac{2\pi}{3}\right) - \sec\left(\frac{2\pi}{3}\right) = -\sqrt{3} - (-2) = 2-\sqrt{3}$
$\tan\left(\frac{\pi}{3}\right) - \sec\left(\frac{\pi}{3}\right) = \sqrt{3} - 2$

$2I = \pi \left[ (2-\sqrt{3}) - (\sqrt{3}-2) \right] = \pi(4 - 2\sqrt{3}) = 2\pi(2-\sqrt{3})$

Thus $I = \pi(2-\sqrt{3})$, which corresponds to option B.