The value of the integral $$\int_0^{\pi/4} \frac{x\,dx}{\sin^4 2x + \cos^4 2x}$$ equals:

We need to evaluate $$I = \int_0^{\pi/4} \frac{x\,dx}{\sin^4 2x + \cos^4 2x}$$.

First, we apply the substitution $$x \to \frac{\pi}{4} - x$$, which gives $$I = \int_0^{\pi/4} \frac{(\pi/4 - x)\,dx}{\sin^4(2(\pi/4 - x)) + \cos^4(2(\pi/4 - x))}.$$

Noting that $$2(\pi/4 - x) = \pi/2 - 2x$$, we have $$\sin(2(\pi/4 - x)) = \cos 2x$$ and $$\cos(2(\pi/4 - x)) = \sin 2x$$, so the integral becomes $$I = \int_0^{\pi/4} \frac{(\pi/4 - x)\,dx}{\cos^4 2x + \sin^4 2x}.$$

By adding the two expressions for $$I$$, we obtain $$2I = \int_0^{\pi/4} \frac{\pi/4}{\sin^4 2x + \cos^4 2x}\,dx$$ and hence $$I = \frac{\pi}{8} \int_0^{\pi/4} \frac{dx}{\sin^4 2x + \cos^4 2x}.$$

Next, we observe that $$\sin^4 2x + \cos^4 2x = (\sin^2 2x + \cos^2 2x)^2 - 2\sin^2 2x\cos^2 2x = 1 - \frac{\sin^2 4x}{2}.$$

Since $$\sin^2 4x = \frac{1 - \cos 8x}{2}$$, it follows that $$\sin^4 2x + \cos^4 2x = 1 - \frac{1 - \cos 8x}{4} = \frac{3 + \cos 8x}{4},$$ and therefore $$I = \frac{\pi}{8} \int_0^{\pi/4} \frac{4\,dx}{3 + \cos 8x}.$$

Setting $$u = 8x$$ so that $$du = 8\,dx$$ and noting that when $$x=0$$, $$u=0$$ and when $$x=\pi/4$$, $$u=2\pi$$, we get $$I = \frac{\pi}{8} \cdot 4 \cdot \frac{1}{8} \int_0^{2\pi} \frac{du}{3 + \cos u} = \frac{\pi}{16} \int_0^{2\pi} \frac{du}{3 + \cos u}.$$

Using the standard formula $$\int_0^{2\pi} \frac{du}{a + b\cos u} = \frac{2\pi}{\sqrt{a^2 - b^2}} \quad (a>b>0),$$ with $$a=3$$ and $$b=1$$ yields $$\int_0^{2\pi} \frac{du}{3 + \cos u} = \frac{2\pi}{\sqrt{9 - 1}} = \frac{\pi}{\sqrt{2}}.$$

Hence, $$I = \frac{\pi}{16} \cdot \frac{\pi}{\sqrt{2}} = \frac{\pi^2}{16\sqrt{2}} = \frac{\sqrt{2}\pi^2}{32},$$ so the correct answer is $$\frac{\sqrt{2}\pi^2}{32}$$ (Option C).