Prove that:
$$\int_{0}^{1}\ln (1+x^2) \arctan x \, {\rm d}x = \ln 2 - \frac{\pi}{2} - \frac{1}{4}\ln^2 2 + \frac{\pi^2}{16} + \frac{\pi \ln 2}{4}$$
Solution
$$\int_{0}^{1}\ln (1+x^2) \arctan x \, {\rm d}x = \ln 2 - \frac{\pi}{2} - \frac{1}{4}\ln^2 2 + \frac{\pi^2}{16} + \frac{\pi \ln 2}{4}$$
Solution