We have, $I=\underset{0}{\overset{\frac{1}{2}}{\int }}\frac{\ln \left(1+2x\right)}{1+4x^2}dx$

Let, $2x=\tan \theta \Rightarrow dx=\frac{1}{2}se{c}^2\theta d\theta$

$\Rightarrow I=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{\ln \left(1+\tan \theta \right)}{se{c}^2\theta }·\frac{1}{2}·se{c}^2\theta d\theta$  $\left(\because 1+{\tan }^2\theta ={\sec }^2\theta \right)$
 $=\frac{1}{2}\underset{0}{\overset{\frac{\pi }{4}}{\int }}ln\left(1+\tan \theta \right)d\theta$
  $=\frac{1}{2}\underset{0}{\overset{\frac{\pi }{4}}{\int }}ln\left(1+\tan \left(\frac{\pi }{4}-\theta \right)\right)d\theta$ 

$\left(\because {\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f\left(a-x\right)dx\right)$$=\frac{1}{2}\underset{0}{\overset{\frac{\pi }{4}}{\int }}ln\left(1+\frac{1-\tan \theta }{1+\tan \theta }\right)d\theta$

 $=\frac{1}{2}\underset{0}{\overset{\frac{\pi }{4}}{\int }}ln\left(\frac{2}{1+\tan \theta }\right)d\theta$$=\frac{1}{2}\left[\left(\ln 2\right)\underset{0}{\overset{\frac{\pi }{4}}{\int }}d\theta -\underset{0}{\overset{\frac{\pi }{4}}{\int }}\ln \left(1+\tan \theta \right)d\theta \right]$
$\Rightarrow 2I=\left(\ln 2\right)\frac{\pi }{4}-2I$

$\Rightarrow I=\frac{\pi }{16}\left(\ln 2\right)$