The value of $$\int\limits_0^1 {\frac{{8\log \left( {1 + x} \right)}}{{1 + {x^2}}}} dx,$$    is-

Solution :
$$\eqalign{ & I = \int\limits_0^1 {\frac{{8\log \left( {1 + x} \right)}}{{1 + {x^2}}}} dx \cr & {\text{Put}}\,x = \tan \,\theta , \cr & \therefore \frac{{dx}}{{d\theta }} = {\sec ^2}\theta \Rightarrow dx = {\sec ^2}\theta \,d\theta \cr & \therefore I = 8\int\limits_0^{\frac{\pi }{4}} {\frac{{\log \left( {1 + \tan \,\theta } \right)}}{{1 + {{\tan }^2}\theta }}} .{\sec ^2}\theta \,d\theta \cr & I = 8\int\limits_0^{\frac{\pi }{4}} {\log \left( {1 + \tan \,\theta } \right)} \,d\theta .....{\text{(i)}} \cr & = 8\int\limits_0^{\frac{\pi }{4}} {\log \left[ {1 + \tan \,\left( {\frac{\pi }{4} - \theta } \right)} \right]} d\theta \cr & = 8\int\limits_0^{\frac{\pi }{4}} {\log \left[ {1 + \frac{{1 - \tan \,\theta }}{{1 + \tan \,\theta }}} \right]} d\theta \cr & = 8\int\limits_0^{\frac{\pi }{4}} {\log \left[ {\frac{2}{{1 + \tan \,\theta }}} \right]d\theta } \cr & = 8\int\limits_0^{\frac{\pi }{4}} {\left[ {\log 2 - \log \left( {1 + \tan \,\theta } \right)} \right]d\theta } \cr & I = 8.\left( {\log \,2} \right)\left[ x \right]_0^{\frac{\pi }{4}} - 8\int\limits_0^{\frac{\pi }{4}} {\log \left( {1 + \tan \,\theta } \right)d\theta } \cr & I = 8.\frac{\pi }{4}.\log \,2 - I\,\,\,\,\left[ {{\text{from equation (i)}}} \right] \cr & \Rightarrow 2I = 2\pi \,\log \,2, \cr & \therefore I = \pi \,\log \,2 \cr} $$