The value of the integral $$\int_{-\pi/4}^{\pi/4} \frac{x + \frac{\pi}{4}}{2 - \cos 2x} dx$$ is :

The integral to evaluate is: $$\int_{-\pi/4}^{\pi/4} \frac{x + \frac{\pi}{4}}{2 - \cos 2x} dx$$

Split the integral into two parts: $$I = \int_{-\pi/4}^{\pi/4} \frac{x}{2 - \cos 2x} dx + \int_{-\pi/4}^{\pi/4} \frac{\frac{\pi}{4}}{2 - \cos 2x} dx$$

Consider the first integral: $$\int_{-\pi/4}^{\pi/4} \frac{x}{2 - \cos 2x} dx$$
The function $$f(x) = \frac{x}{2 - \cos 2x}$$ is odd because $$f(-x) = \frac{-x}{2 - \cos(-2x)} = \frac{-x}{2 - \cos 2x} = -f(x)$$.
Since the limits are symmetric about zero, the integral of an odd function over a symmetric interval is zero. Thus: $$\int_{-\pi/4}^{\pi/4} \frac{x}{2 - \cos 2x} dx = 0$$

Therefore: $$I = 0 + \int_{-\pi/4}^{\pi/4} \frac{\frac{\pi}{4}}{2 - \cos 2x} dx = \frac{\pi}{4} \int_{-\pi/4}^{\pi/4} \frac{1}{2 - \cos 2x} dx$$

Now, consider the function $$g(x) = \frac{1}{2 - \cos 2x}$$. It is even because $$g(-x) = \frac{1}{2 - \cos(-2x)} = \frac{1}{2 - \cos 2x} = g(x)$$.
Since the function is even and the limits are symmetric: $$\int_{-\pi/4}^{\pi/4} \frac{1}{2 - \cos 2x} dx = 2 \int_{0}^{\pi/4} \frac{1}{2 - \cos 2x} dx$$

Thus: $$I = \frac{\pi}{4} \times 2 \int_{0}^{\pi/4} \frac{1}{2 - \cos 2x} dx = \frac{\pi}{2} \int_{0}^{\pi/4} \frac{1}{2 - \cos 2x} dx$$

Substitute $$t = 2x$$, so $$dt = 2 dx$$ or $$dx = \frac{dt}{2}$$.
When $$x = 0$$, $$t = 0$$; when $$x = \frac{\pi}{4}$$, $$t = \frac{\pi}{2}$$.
The integral becomes: $$\int_{0}^{\pi/4} \frac{1}{2 - \cos 2x} dx = \int_{0}^{\pi/2} \frac{1}{2 - \cos t} \cdot \frac{dt}{2} = \frac{1}{2} \int_{0}^{\pi/2} \frac{1}{2 - \cos t} dt$$

Therefore: $$I = \frac{\pi}{2} \times \frac{1}{2} \int_{0}^{\pi/2} \frac{1}{2 - \cos t} dt = \frac{\pi}{4} \int_{0}^{\pi/2} \frac{1}{2 - \cos t} dt$$

Evaluate $$\int_{0}^{\pi/2} \frac{1}{2 - \cos t} dt$$ using the substitution $$u = \tan \frac{t}{2}$$.
Then, $$\cos t = \frac{1 - u^2}{1 + u^2}$$ and $$dt = \frac{2 du}{1 + u^2}$$.
When $$t = 0$$, $$u = \tan 0 = 0$$; when $$t = \frac{\pi}{2}$$, $$u = \tan \frac{\pi}{4} = 1$$.

The integral becomes: $$\int_{0}^{\pi/2} \frac{1}{2 - \cos t} dt = \int_{0}^{1} \frac{1}{2 - \frac{1 - u^2}{1 + u^2}} \cdot \frac{2 du}{1 + u^2}$$

Simplify the denominator: $$2 - \frac{1 - u^2}{1 + u^2} = \frac{2(1 + u^2) - (1 - u^2)}{1 + u^2} = \frac{2 + 2u^2 - 1 + u^2}{1 + u^2} = \frac{1 + 3u^2}{1 + u^2}$$

So: $$\int_{0}^{1} \frac{1}{\frac{1 + 3u^2}{1 + u^2}} \cdot \frac{2 du}{1 + u^2} = \int_{0}^{1} \frac{1 + u^2}{1 + 3u^2} \cdot \frac{2 du}{1 + u^2} = \int_{0}^{1} \frac{2}{1 + 3u^2} du = 2 \int_{0}^{1} \frac{1}{1 + 3u^2} du$$

Rewrite as: $$2 \int_{0}^{1} \frac{1}{1 + (\sqrt{3} u)^2} du$$

Substitute $$v = \sqrt{3} u$$, so $$dv = \sqrt{3} du$$ and $$du = \frac{dv}{\sqrt{3}}$$.
When $$u = 0$$, $$v = 0$$; when $$u = 1$$, $$v = \sqrt{3}$$.

Thus: $$2 \int_{0}^{1} \frac{1}{1 + (\sqrt{3} u)^2} du = 2 \int_{0}^{\sqrt{3}} \frac{1}{1 + v^2} \cdot \frac{dv}{\sqrt{3}} = \frac{2}{\sqrt{3}} \int_{0}^{\sqrt{3}} \frac{1}{1 + v^2} dv$$

The integral of $$\frac{1}{1 + v^2}$$ is $$\tan^{-1} v$$: $$\frac{2}{\sqrt{3}} \left[ \tan^{-1} v \right]_{0}^{\sqrt{3}} = \frac{2}{\sqrt{3}} \left( \tan^{-1} \sqrt{3} - \tan^{-1} 0 \right) = \frac{2}{\sqrt{3}} \left( \frac{\pi}{3} - 0 \right) = \frac{2}{\sqrt{3}} \cdot \frac{\pi}{3} = \frac{2\pi}{3\sqrt{3}}$$

Therefore: $$\int_{0}^{\pi/2} \frac{1}{2 - \cos t} dt = \frac{2\pi}{3\sqrt{3}}$$

Substitute back: $$I = \frac{\pi}{4} \times \frac{2\pi}{3\sqrt{3}} = \frac{\pi}{4} \cdot \frac{2\pi}{3\sqrt{3}} = \frac{2\pi^2}{12\sqrt{3}} = \frac{\pi^2}{6\sqrt{3}}$$

The value of the integral is $$\frac{\pi^2}{6\sqrt{3}}$$, which corresponds to option D.