Integral: $\int_0^{1/2}\frac{\ln(1+2x)}{1+4x^2}{\rm d}x$ [Other method type question.] 
Integral: $$I=\int_0^{1/2}\frac{\ln(1+2x)}{1+4x^2}{\rm d}x$$

One thing to quickly do it take $y=2x$:
$$I=\frac12\int_0^1\frac{\ln(1+y)}{1+y^2}{\rm d}y$$
I took $y=\tan z$:
$$I=\frac12\int_0^{\pi/4}\ln(1+\tan z){\rm d}z$$
Now substitute $u=\pi/4-z$:
$$I=\frac12\int_0^{\pi/4}[\ln2-\ln(1+\tan z)]{\rm d}z=\frac\pi8\ln2-I\\I=\frac\pi{16}\ln 2$$
Any other methods?
 A: We have
\begin{align}I &= \frac{1}{2} \int_0^1 \frac{\ln(1 + y)}{1 + y^2}\, dy\\
& = \frac{1}{2}\int_0^1\int_0^1 \frac{y}{(1 + ry)(1 + y^2)}\, dr\, dy\\
& = \frac{1}{2}\int_0^1 \int_0^1 \frac{y}{(1 + ry)(1 + y^2)}\, dy\, dr\\
&= \frac{1}{2}\int_0^1 \int_0^1 \left(\frac{r}{(1 + r^2)(1 + y^2)} + \frac{y}{(1 + r^2)(1 + y^2)}- \frac{r}{(1 + r^2)(1 + ry)}\right)\, dy\, dr\\
&= \frac{1}{2}\int_0^1 \left(\frac{r}{1 + r^2}\cdot \frac{\pi}{4} + \frac{1}{2(1 + r^2)}\ln(2) - \frac{\ln(1 + r)}{1 + r^2}\right)\, dr\\
&= \frac{1}{2}\left(\frac{\pi}{8}\ln(2) + \frac{\pi}{8}\ln(2)\right) - I\\
&= \frac{\pi}{8}\ln(2) - I
\end{align}
So $2I = \frac{\pi}{8}\ln(2)$, or $$I = \frac{\pi}{16}\ln(2).$$
A: Once we have:
$$ I = \frac{1}{2}\int_{0}^{1}\frac{\log(1+y)}{1+y^2}\,dy$$
we can set $y=\frac{z-1}{z+1}$ in order to have:
$$ I = \int_{1}^{+\infty}\frac{\log\frac{2z}{z+1}}{(z-1)^2+(z+1)^2}\,dz=\int_{0}^{1}\frac{\log\frac{2}{z+1}}{2z^2+2}\,dz=\frac{\pi}{8}\log 2-\frac{1}{2}\int_{0}^{1}\frac{\log(z+1)}{z^2+1}\,dz$$
or:
$$ 2I = \frac{\pi}{8}\log 2\implies I = \frac{\pi}{16}\log 2$$
just as you got.
