Answer to Question #166449 in Calculus for Phyroe

Question

∫(tan^-1 x)/(1+x² )dx from 0 to 1

Answer 1

$\displaystyle\int_0^1\frac{\tan^{-1}x}{1+x^2}\,dx \\ = \int_0^1\tan^{-1}x\,\frac d{dx}(\tan^{-1}x)\,dx \\ = \frac12(\tan^{-1}x)^2\Big|_0^1 \\ = \frac12[(\tan^{-1}1)^2 - \tan^{-1}0)^2] \\ = \frac12[(\frac \pi4)^2 - 0] \\ = \frac {\pi^2}{32}$