### $I = \int\limits_0^1 {\frac{{\ln (1 + x)}}{{1 + {x^2}}}dx} = ?$

Let, $x = \tan \theta$

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

$I = \int\limits_0^{\pi /4} {\frac{{\ln (1 + \tan \theta )}}{{1 + {{\tan }^2}\theta }}{{\sec }^2}\theta d\theta }$

$= \int\limits_0^{\pi /4} {\ln \left( {\frac{{\sin \theta + \cos \theta }}{{\cos \theta }}} \right)d\theta }$

$= \int\limits_0^{\pi /4} {\ln (\sin \theta + \cos \theta )d\theta - \int\limits_0^{\pi /4} {\ln \cos \theta d\theta } }$

$= \int\limits_0^{\pi /4} {\ln \left\{ {\sqrt 2 \left( {\cos \theta .\frac{1}{{\sqrt 2 }} + \sin \theta .\frac{1}{{\sqrt 2 }}} \right)} \right\}d\theta - \int\limits_0^{\pi /4} {\ln \cos \theta d\theta } }$

$= \int\limits_0^{\pi /4} {\ln \left\{ {\sqrt 2 \left( {\cos \theta .\cos \frac{\pi }{4} + \sin \theta .\sin \frac{\pi }{4}} \right)} \right\}d\theta - \int\limits_0^{\pi /4} {\ln \cos \theta d\theta } }$

$= \int\limits_0^{\pi /4} {\ln \left\{ {\sqrt 2 \cos \left( {\theta - \frac{\pi }{4}} \right)} \right\}d\theta - \int\limits_0^{\pi /4} {\ln \cos \theta d\theta } }$

$= \int\limits_0^{\pi /4} {\ln \sqrt 2 d\theta + } \int\limits_0^{\pi /4} {\ln \cos \left( {\theta - \frac{\pi }{4}} \right)d\theta - \int\limits_0^{\pi /4} {\ln \cos \theta d\theta } }$

Let, $\theta - \frac{\pi }{4} = - \phi$ for the second integral.

$I = \frac{\pi }{4}\ln \sqrt 2 - \int\limits_{\pi /4}^0 {\ln \cos ( - \phi )d\phi - \int\limits_0^{\pi /4} {\ln \cos \theta d\theta } }$

$= \frac{\pi }{4}\ln \sqrt 2 + \int\limits_0^{\pi /4} {\ln \cos \phi d\phi - \int\limits_0^{\pi /4} {\ln \cos \theta d\theta } }$

$= \frac{\pi }{4}\ln \sqrt 2 = \frac{\pi }{8}\ln 2$