Answer
$$
\int_{0}^{\pi / 8} \arctan 2 x d x=\frac{\pi}{8} \arctan \frac{\pi}{4}-\frac{1}{4} \ln \left(1+\frac{\pi^{2}}{16}\right)
$$
Work Step by Step
$$
\int_{0}^{\pi / 8} \arctan 2 x d x
$$
Let
$$
u=2 x \quad d u=2d x
$$
and at
$$ x=\pi / 8 \rightarrow u=\pi / 4, \quad x=0 \rightarrow u=0
$$
If we look at the Table of Integrals , we see that the closest entry is number $89$ :
$$
\begin{aligned}
\int_{0}^{\pi / 8} \arctan 2 x d x &=\frac{1}{2} \int_{0}^{\pi / 4} \arctan u d u \quad[u=2 x, d u=2 d x] \\
& \stackrel{89}{=} \frac{1}{2}\left[u \arctan u-\frac{1}{2} \ln \left(1+u^{2}\right)\right]_{0}^{\pi / 4}=\frac{1}{2}\left\{\left[\frac{\pi}{4} \arctan \frac{\pi}{4}-\frac{1}{2} \ln \left(1+\frac{\pi^{2}}{16}\right)\right]-0\right\} \\
&=\frac{\pi}{8} \arctan \frac{\pi}{4}-\frac{1}{4} \ln \left(1+\frac{\pi^{2}}{16}\right)
\end{aligned}
$$
