The value of the integral $$\int_{-\pi/2}^{\pi/2} \sin^4 x\left(1 + \log\left(\frac{2+\sin x}{2-\sin x}\right)\right)dx$$ is:

We need to evaluate $$I = \int_{-\pi/2}^{\pi/2} \sin^4 x\left(1 + \log\left(\frac{2+\sin x}{2-\sin x}\right)\right)dx$$.

Expanding the integrand, we get $$I = \int_{-\pi/2}^{\pi/2} \sin^4 x\, dx + \int_{-\pi/2}^{\pi/2} \sin^4 x \cdot \log\left(\frac{2+\sin x}{2-\sin x}\right)dx$$.

Let $$I_1 = \int_{-\pi/2}^{\pi/2} \sin^4 x\, dx$$ and $$I_2 = \int_{-\pi/2}^{\pi/2} \sin^4 x \cdot \log\left(\frac{2+\sin x}{2-\sin x}\right)dx$$.

For $$I_2$$, define $$g(x) = \sin^4 x \cdot \log\left(\frac{2+\sin x}{2-\sin x}\right)$$. We check $$g(-x) = \sin^4(-x) \cdot \log\left(\frac{2+\sin(-x)}{2-\sin(-x)}\right) = \sin^4 x \cdot \log\left(\frac{2-\sin x}{2+\sin x}\right)$$.

Since $$\log\left(\frac{2-\sin x}{2+\sin x}\right) = -\log\left(\frac{2+\sin x}{2-\sin x}\right)$$, we have $$g(-x) = -g(x)$$. So $$g(x)$$ is an odd function.

The integral of an odd function over a symmetric interval $$[-a, a]$$ is zero. Therefore $$I_2 = 0$$.

For $$I_1$$, since $$\sin^4 x$$ is even, $$I_1 = 2\int_0^{\pi/2} \sin^4 x\, dx$$.

Using the reduction formula: $$\int_0^{\pi/2} \sin^4 x\, dx = \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} = \frac{3\pi}{16}$$.

Therefore $$I_1 = 2 \cdot \frac{3\pi}{16} = \frac{3\pi}{8}$$.

The value of the integral is $$I = I_1 + I_2 = \frac{3\pi}{8} + 0 = \frac{3\pi}{8}$$.

The correct answer is Option B: $$\frac{3}{8}\pi$$.