Question

Evaluate ${\int }_0^1\frac{\log (1+x)}{1+x^2}dx$

Answer 1

${\int }_0^1\frac{\log (1+n)}{1+n^2}dx$

Put $n=\tan \theta$ hence $dx={\sec }^2\theta d\theta$ and limits get changed to $0$ to $\pi /4$

${\int }_0^{\pi /4}\frac{\log (1+\tan \theta )}{1+{\tan }^2\theta }{\sec }^2\theta d\theta$

and $1+{\tan }^2\theta ={\sec }^2\theta$

${\int }_0^{\pi /4}\frac{\log (1+\tan \theta )}{{\sec }^2\theta }{\sec }^2\theta d\theta$

$I={\int }_0^{\pi /4}\log (1+\tan \theta )d\theta (\therefore {\int }_0^{a}f\left(x\right)dx,{\int }_0^{a}f(a-x)dx)$

we get

$I={\int }_0^{\pi /4}\log \left(2\right)d\theta$

$I=\left(\log \left(2\right)\right)\frac{\pi }{4}$

$I=\frac{\pi }{8}\log \left(2\right)$