Answer
$$\int_{0}^{\pi/8}arctan\,2x\,dx=\frac{\pi}{8}arctan\,\frac{\pi}{4}-\frac{1}{4}ln(1+\frac{\pi^{2}}{16})$$
Work Step by Step
Table of integrals 89:
$\int tan^{-1}u\,du=u\,tan^{-1}u-\frac{1}{2}ln(1+u^{2})+C$
So$$\int_{0}^{\pi/8}arctan\,2x\,dx=\frac{1}{2}\int_{0}^{\pi/4}arctanx\,dx$$
$$=\frac{1}{2}\left [ x\,tan^{-1}x-\frac{1}{2}ln(1+x^{2}) \right ]\int_{0}^{\pi/4}$$
$$=\frac{\pi}{8}arctan\,\frac{\pi}{4}-\frac{1}{4}ln(1+\frac{\pi^{2}}{16})$$
