Question

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

Answer 1

$I={\int }_0^1\frac{log(1+x)}{1+x^2}dx$
$\begin{array}{ccc}x=tan\theta & x=0& tan\theta =0\\ & & \theta =0\\ dx=\frac{1}{1+x^2}& x=1& tan\theta =1\\ & & \theta =ta{n}^{-1}1=\frac{\pi }{4}\end{array}$
$I={\int }_0^{\frac{\pi }{4}}log(1+tan\theta )d\theta$ $\to \left(1\right)$
${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx$
$I={\int }_0^{\frac{\pi }{4}}log(1+\frac{1-tan\theta }{1+tan\theta })d\theta$
$={\int }_0^{\frac{\pi }{4}}log\frac{2}{1+tan\theta }d\theta$ $\to \left(2\right)$
(1) + (2)
$I={\int }_0^1\frac{log(1+x)}{1+x^2}dx$
$2I={\int }_0^{\frac{\pi }{4}}log(1+tan\theta )+log2-log(1+tan\theta )d\theta$
$2I={\int }_0^{\frac{\pi }{4}}log2d\theta$
$I=\frac{log2}{2}\theta {\int }_0^{\frac{\pi }{4}}$
$=lot2[\frac{\pi }{8}-\theta ]$
$=\frac{\pi }{8}log2$