If $$\int \frac{1}{x}\sqrt{\frac{1-x}{1+x}} dx = g(x) + c, g(1) = 0$$, then $$g\left(\frac{1}{2}\right)$$ is equal to

We need to evaluate $$\int \frac{1}{x}\sqrt{\frac{1-x}{1+x}}\,dx$$.

Let $$x = \cos\theta$$, so $$dx = -\sin\theta\,d\theta$$.

Then $$\sqrt{\frac{1-x}{1+x}} = \sqrt{\frac{1-\cos\theta}{1+\cos\theta}} = \sqrt{\frac{2\sin^2(\theta/2)}{2\cos^2(\theta/2)}} = \tan\frac{\theta}{2}$$.

Substituting these expressions into the integral gives $$\int \frac{1}{\cos\theta}\cdot\tan\frac{\theta}{2}\cdot(-\sin\theta)\,d\theta$$.

Since $$\sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$$, this becomes $$-\int \frac{\tan(\theta/2)\cdot 2\sin(\theta/2)\cos(\theta/2)}{\cos\theta}\,d\theta = -\int \frac{2\sin^2(\theta/2)}{\cos\theta}\,d\theta$$.

Using $$2\sin^2(\theta/2) = 1 - \cos\theta$$ transforms the integral to $$-\int \frac{1-\cos\theta}{\cos\theta}\,d\theta = -\int(\sec\theta - 1)\,d\theta$$.

Integrating yields $$-\ln|\sec\theta + \tan\theta| + \theta + C$$.

Since $$x = \cos\theta$$, it follows that $$\theta = \cos^{-1}x$$, $$\sec\theta = \frac{1}{x}$$, and $$\tan\theta = \frac{\sqrt{1-x^2}}{x}$$.

Hence $$g(x) = -\ln\left|\frac{1}{x} + \frac{\sqrt{1-x^2}}{x}\right| + \cos^{-1}x = -\ln\left(\frac{1+\sqrt{1-x^2}}{x}\right) + \cos^{-1}x$$.

Applying the condition $$g(1) = 0$$ gives $$g(1) = -\ln\left(\frac{1+0}{1}\right) + \cos^{-1}(1) = -\ln 1 + 0 = 0$$, so $$C = 0$$.

To compute $$g\left(\frac{1}{2}\right)$$, substitute into the expression: $$g\left(\frac{1}{2}\right) = -\ln\left(\frac{1+\sqrt{1-1/4}}{1/2}\right) + \cos^{-1}\left(\frac{1}{2}\right)$$.

It follows that $$g\left(\frac{1}{2}\right) = -\ln\left(\frac{1+\frac{\sqrt{3}}{2}}{1/2}\right) + \frac{\pi}{3}$$.

Now, $$2+\sqrt{3} = \frac{(2+\sqrt{3})(\sqrt{3}+1)}{(\sqrt{3}+1)} \cdot \frac{1}{1}$$. More directly, note that $$\frac{\sqrt{3}+1}{\sqrt{3}-1} = \frac{(\sqrt{3}+1)^2}{2} = \frac{4+2\sqrt{3}}{2} = 2+\sqrt{3}$$.

Therefore:

$$g\left(\frac{1}{2}\right) = -\ln\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right) + \frac{\pi}{3} = \ln\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right) + \frac{\pi}{3}$$.

Hence the correct answer is Option A: $$\log_e\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right) + \frac{\pi}{3}$$.