Question

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

Answer 1

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

Put $x=\tan$

$\therefore dx={\sec }^2\theta d\theta$

limits are

$I={\int }_0^{\pi /4}\log (1+\tan \theta )d\theta ......\left(1\right)$

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

Now we know that ${\int }_0^{a}f\left(x\right)dx=\int f(a-x)dx$

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

and $\tan (\frac{\pi }{4}-\theta )=\frac{\tan \frac{\pi }{4}-\tan \theta }{1+\tan \frac{\pi }{4}\tan \theta }=\frac{1-\tan \theta }{1+\tan \theta }$

$\therefore I={\int }_0^{\pi /4}\log (1+\frac{1-\tan \theta }{1+\tan \theta })d\theta$

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

$I={\int }_0^{\pi /4}[\log 2-\log (1+\tan \theta \left)\right]d\theta$

$I={\int }_0^{\pi /4}\log 2d\theta -I$

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

$I=\frac{1}{2}\log 2[0{]}_0^{\pi /4}$

$I=\frac{1}{2}\log 2[\frac{\pi }{4}-0]$

$I=\frac{\pi }{8}\log 2$