The value of the integral $$\int_{1/2}^{2} \frac{\tan^{-1}x}{x} dx$$ is equal to

We need to evaluate $$\int_{1/2}^{2} \frac{\tan^{-1}x}{x} dx$$.

Using the substitution $$x = \frac{1}{t}$$ and $$dx = -\frac{1}{t^2}dt$$, we find that when $$x = 1/2$$, $$t = 2$$, and when $$x = 2$$, $$t = 1/2$$. Therefore,

$$I = \int_{1/2}^{2} \frac{\tan^{-1}x}{x}dx = \int_2^{1/2} \frac{\tan^{-1}(1/t)}{1/t}\cdot\left(-\frac{1}{t^2}\right)dt = \int_{1/2}^{2} \frac{\tan^{-1}(1/t)}{t}dt$$

Using $$\tan^{-1}(1/t) = \frac{\pi}{2} - \tan^{-1}t$$ for $$t > 0$$ gives

$$I = \int_{1/2}^{2} \frac{\frac{\pi}{2} - \tan^{-1}t}{t}dt = \frac{\pi}{2}\int_{1/2}^{2}\frac{dt}{t} - \int_{1/2}^{2}\frac{\tan^{-1}t}{t}dt$$

It follows that

$$I = \frac{\pi}{2}\ln\frac{2}{1/2} - I = \frac{\pi}{2}\ln 4 - I$$

Hence,

$$2I = \frac{\pi}{2}\cdot 2\ln 2 = \pi\ln 2$$

and thus

$$I = \frac{\pi}{2}\ln 2$$

The answer is $$\boxed{\frac{\pi}{2}\log_e 2}$$, which is Option D.