Question

I = ${\int }_0^1\frac{\log \left(1+x\right)}{1+x^2}dx$

Answer 1

$\begin{array}{c}{\int }_0^1\frac{\log \left(1+x\right)}{1+x^2}dx\\ Letx=\tan \theta \\ {\sec }^2\theta \propto \theta =dx\\ I={\int }_0^{\pi /2}\frac{\log \left(1+\tan \theta \right)}{{\sec }^2\theta }\left({\sec }^2\theta d\theta \right)\\ I={\int }_0^{\pi /4}\log \left(1+\tan \theta \right)d\theta \_\_\_\_\_\left(1\right)\\ I={\int }_0^{\pi /4}\log \left(1+\frac{1-\tan \theta }{1+\tan \theta }\right)d\theta \\ I={\int }_0^{\pi /4}\log \left(\frac{2}{1+\tan \theta }\right)d\theta \_\_\_\_\_\left(2\right)\\ \left(1\right)+\left(2\right)\_\_\_2I={\int }_0^{\pi /4}\log 2d\theta \\ I=\frac{1}{2}{{\left[\theta \log 2\right]}_0}^{\pi /4}\\ I=\frac{\pi }{8}\log 2\end{array}$