The integral $$\int_0^{\pi/2} \frac{1}{3 + 2\sin x + \cos x} dx$$ is equal to:

We need to evaluate $$\displaystyle\int_0^{\pi/2} \frac{dx}{3 + 2\sin x + \cos x}$$.

We use the Weierstrass substitution $$t = \tan\!\left(\dfrac{x}{2}\right)$$, which gives $$\sin x = \dfrac{2t}{1 + t^2}$$, $$\cos x = \dfrac{1 - t^2}{1 + t^2}$$, and $$dx = \dfrac{2\,dt}{1 + t^2}$$. When $$x = 0$$, $$t = 0$$; when $$x = \pi/2$$, $$t = 1$$.

The denominator becomes $$3 + \dfrac{4t}{1 + t^2} + \dfrac{1 - t^2}{1 + t^2} = \dfrac{3(1 + t^2) + 4t + 1 - t^2}{1 + t^2} = \dfrac{2t^2 + 4t + 4}{1 + t^2} = \dfrac{2(t^2 + 2t + 2)}{1 + t^2}$$.

Hence the integral transforms to $$\displaystyle\int_0^1 \frac{1}{\dfrac{2(t^2 + 2t + 2)}{1 + t^2}} \cdot \frac{2\,dt}{1 + t^2} = \int_0^1 \frac{(1 + t^2)}{2(t^2 + 2t + 2)} \cdot \frac{2\,dt}{1 + t^2} = \int_0^1 \frac{dt}{t^2 + 2t + 2}$$.

Now we complete the square: $$t^2 + 2t + 2 = (t + 1)^2 + 1$$. Substituting $$u = t + 1$$, $$du = dt$$, with limits $$u = 1$$ to $$u = 2$$:

$$\displaystyle\int_1^2 \frac{du}{u^2 + 1} = \Big[\tan^{-1}(u)\Big]_1^2 = \tan^{-1}(2) - \tan^{-1}(1) = \tan^{-1}(2) - \frac{\pi}{4}$$.

Hence, the correct answer is Option B.