The value of $$\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \dfrac{2 + 3\sin x}{\sin x(1 + \cos x)} dx$$ is equal to

We need to evaluate $$I = \int_{\pi/3}^{\pi/2} \frac{2 + 3\sin x}{\sin x(1 + \cos x)}\, dx$$. To simplify this integral, we employ the Weierstrass substitution by setting $$t = \tan(x/2)$$, which yields the identities $$\sin x = \frac{2t}{1+t^2}$$, $$\cos x = \frac{1-t^2}{1+t^2}$$ and $$dx = \frac{2\,dt}{1+t^2}$$. As a result, $$1 + \cos x = \frac{2}{1+t^2}$$, and when $$x = \pi/3$$, we have $$t = \tan(\pi/6) = 1/\sqrt{3}$$ while for $$x = \pi/2$$, $$t = \tan(\pi/4) = 1$$.

Substituting these expressions into the integral transforms it into $$I = \int_{1/\sqrt{3}}^{1} \frac{2 + \frac{6t}{1+t^2}}{\frac{2t}{1+t^2} \cdot \frac{2}{1+t^2}} \cdot \frac{2\,dt}{1+t^2}\,. $$ The numerator simplifies as $$2 + \frac{6t}{1+t^2} = \frac{2+2t^2+6t}{1+t^2} = \frac{2(t^2+3t+1)}{1+t^2}$$ while the denominator becomes $$\frac{4t}{(1+t^2)^2}\,. $$ Therefore the integrand reduces to $$\frac{2(t^2+3t+1)}{1+t^2} \cdot \frac{(1+t^2)^2}{4t} \cdot \frac{2}{1+t^2} = \frac{t^2+3t+1}{t}\,, $$ and the integral takes the form $$I = \int_{1/\sqrt{3}}^{1}\left(t + 3 + \frac{1}{t}\right) dt\,. $$

Integrating term by term gives $$\int\left(t + 3 + \frac{1}{t}\right) dt = \frac{t^2}{2} + 3t + \ln t\,, $$ so that $$I = \left[\frac{t^2}{2} + 3t + \ln t\right]_{1/\sqrt{3}}^{1}\,. $$ At the upper limit $$t = 1$$ this expression equals $$\frac{1}{2} + 3 + 0 = \frac{7}{2}$$, and at the lower limit $$t = 1/\sqrt{3}$$ it becomes $$\frac{1}{6} + \frac{3}{\sqrt{3}} + \ln(1/\sqrt{3}) = \frac{1}{6} + \sqrt{3} - \frac{1}{2}\ln 3\,. $$ Subtracting yields $$I = \frac{7}{2} - \left(\frac{1}{6} + \sqrt{3} - \frac{1}{2}\ln 3\right) = \frac{7}{2} - \frac{1}{6} - \sqrt{3} + \frac{1}{2}\ln 3\,. $$ This simplifies further to $$\frac{21-1}{6} - \sqrt{3} + \ln\sqrt{3} = \frac{20}{6} - \sqrt{3} + \ln\sqrt{3} = \frac{10}{3} - \sqrt{3} + \ln\sqrt{3}\,. $$

Hence the correct answer is Option (3): $$\boxed{\frac{10}{3} - \sqrt{3} + \log_e\sqrt{3}}$$.