The value of integral $$\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{x}{1 + \sin x} dx$$ is:

We need to evaluate $$I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{x}{1 + \sin x}\, dx$$.

We use the King's property of definite integrals: $$\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx$$.

Here $$a + b = \frac{\pi}{4} + \frac{3\pi}{4} = \pi$$, so:

$$I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{\pi - x}{1 + \sin(\pi - x)}\, dx = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{\pi - x}{1 + \sin x}\, dx$$

since $$\sin(\pi - x) = \sin x$$.

Adding the two expressions for $$I$$:

$$2I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{x + (\pi - x)}{1 + \sin x}\, dx = \pi \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{1}{1 + \sin x}\, dx$$

To evaluate $$\int \frac{1}{1 + \sin x}\, dx$$, multiply numerator and denominator by $$(1 - \sin x)$$:

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

$$= \sec^2 x - \sec x \tan x$$

So $$\int \frac{1}{1 + \sin x}\, dx = \tan x - \sec x + C$$.

Now evaluate from $$\frac{\pi}{4}$$ to $$\frac{3\pi}{4}$$:

At $$x = \frac{3\pi}{4}$$: $$\tan\frac{3\pi}{4} = -1$$ and $$\sec\frac{3\pi}{4} = -\sqrt{2}$$, so $$\tan\frac{3\pi}{4} - \sec\frac{3\pi}{4} = -1 - (-\sqrt{2}) = \sqrt{2} - 1$$.

At $$x = \frac{\pi}{4}$$: $$\tan\frac{\pi}{4} = 1$$ and $$\sec\frac{\pi}{4} = \sqrt{2}$$, so $$\tan\frac{\pi}{4} - \sec\frac{\pi}{4} = 1 - \sqrt{2}$$.

Therefore: $$\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{1}{1 + \sin x}\, dx = (\sqrt{2} - 1) - (1 - \sqrt{2}) = 2\sqrt{2} - 2 = 2(\sqrt{2} - 1)$$

Substituting back: $$2I = \pi \cdot 2(\sqrt{2} - 1) = 2\pi(\sqrt{2} - 1)$$

$$I = \pi(\sqrt{2} - 1)$$

The answer is Option B: $$\pi(\sqrt{2} - 1)$$.