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

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$