The integral $$\int_0^{\frac{1}{2}} \frac{\ln(1+2x)}{1+4x^2} dx$$ equals:

To solve the integral $$\int_0^{\frac{1}{2}} \frac{\ln(1+2x)}{1+4x^2} dx$$, start by making a substitution to simplify the expression. Set $$u = 2x$$, which gives $$du = 2 dx$$ or $$dx = \frac{du}{2}$$. The limits change as follows: when $$x = 0$$, $$u = 0$$; when $$x = \frac{1}{2}$$, $$u = 1$$. Substitute these into the integral:

$$$ \int_0^{\frac{1}{2}} \frac{\ln(1+2x)}{1+4x^2} dx = \int_0^1 \frac{\ln(1+u)}{1+u^2} \cdot \frac{du}{2} = \frac{1}{2} \int_0^1 \frac{\ln(1+u)}{1+u^2} du $$$

Now, focus on evaluating $$\int_0^1 \frac{\ln(1+u)}{1+u^2} du$$. Use the trigonometric substitution $$u = \tan \theta$$, so $$du = \sec^2 \theta d\theta$$. The limits change: when $$u = 0$$, $$\theta = 0$$; when $$u = 1$$, $$\theta = \frac{\pi}{4}$$. Also, $$1 + u^2 = 1 + \tan^2 \theta = \sec^2 \theta$$. Substitute these into the integral:

$$$ \int_0^1 \frac{\ln(1+u)}{1+u^2} du = \int_0^{\frac{\pi}{4}} \frac{\ln(1 + \tan \theta)}{\sec^2 \theta} \cdot \sec^2 \theta d\theta = \int_0^{\frac{\pi}{4}} \ln(1 + \tan \theta) d\theta $$$

Denote this integral by $$I$$, so:

$$$ I = \int_0^{\frac{\pi}{4}} \ln(1 + \tan \theta) d\theta $$$

To evaluate $$I$$, use the substitution $$\phi = \frac{\pi}{4} - \theta$$. Then $$d\phi = -d\theta$$. The limits change: when $$\theta = 0$$, $$\phi = \frac{\pi}{4}$$; when $$\theta = \frac{\pi}{4}$$, $$\phi = 0$$. Now, express $$\tan \theta$$ in terms of $$\phi$$:

$$$ \tan \theta = \tan\left(\frac{\pi}{4} - \phi\right) = \frac{1 - \tan \phi}{1 + \tan \phi} $$$

So,

$$$ 1 + \tan \theta = 1 + \frac{1 - \tan \phi}{1 + \tan \phi} = \frac{(1 + \tan \phi) + (1 - \tan \phi)}{1 + \tan \phi} = \frac{2}{1 + \tan \phi} $$$

Taking the natural logarithm:

$$$ \ln(1 + \tan \theta) = \ln\left(\frac{2}{1 + \tan \phi}\right) = \ln 2 - \ln(1 + \tan \phi) $$$

Substitute into the integral for $$I$$:

$$$ I = \int_{\frac{\pi}{4}}^{0} \left( \ln 2 - \ln(1 + \tan \phi) \right) (-d\phi) = \int_0^{\frac{\pi}{4}} \left( \ln 2 - \ln(1 + \tan \phi) \right) d\phi $$$

Since $$\phi$$ is a dummy variable, it can be replaced by $$\theta$$:

$$$ I = \int_0^{\frac{\pi}{4}} \ln 2 d\theta - \int_0^{\frac{\pi}{4}} \ln(1 + \tan \theta) d\theta = \ln 2 \cdot \left[ \theta \right]_0^{\frac{\pi}{4}} - I $$$

Evaluate the first integral:

$$$ \int_0^{\frac{\pi}{4}} \ln 2 d\theta = \ln 2 \cdot \left( \frac{\pi}{4} - 0 \right) = \frac{\pi}{4} \ln 2 $$$

So,

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

Solve for $$I$$:

$$$ I + I = \frac{\pi}{4} \ln 2 $$$

$$$ 2I = \frac{\pi}{4} \ln 2 $$$

$$$ I = \frac{\pi}{8} \ln 2 $$$

Thus, $$\int_0^{\frac{\pi}{4}} \ln(1 + \tan \theta) d\theta = \frac{\pi}{8} \ln 2$$, and therefore:

$$$ \int_0^1 \frac{\ln(1+u)}{1+u^2} du = \frac{\pi}{8} \ln 2 $$$

Now substitute back into the original expression:

$$$ \frac{1}{2} \int_0^1 \frac{\ln(1+u)}{1+u^2} du = \frac{1}{2} \cdot \frac{\pi}{8} \ln 2 = \frac{\pi}{16} \ln 2 $$$

Hence, the integral $$\int_0^{\frac{1}{2}} \frac{\ln(1+2x)}{1+4x^2} dx = \frac{\pi}{16} \ln 2$$. Comparing with the options, this matches option B.

So, the answer is Option B.