Mathematics - Definite Integration Question with Solution | TestHub

$\mathrm{I}=8{\int }_0^1\frac{\log (1+\mathrm{x})}{1+{\mathrm{x}}^2}\mathrm{dx}$ $=8{\int }_0^{\frac{\pi }{4}}\frac{\log (1+\tan \theta )}{1+{\tan }^2\theta }{\sec }^2\theta d\theta (\mathrm{let}x=\tan \theta )$ $=8{\int }_0^{\frac{\pi }{4}}\log \left(1+\tan \left(\frac{\pi }{4}-\theta \right)\right)d\theta =8{\int }_0^{\frac{\pi }{4}}\log \left(1+\frac{1-\tan \theta }{1+\tan \theta }\right)d\theta =8{\int }_0^{\frac{\pi }{4}}\log 2d\theta -8{\int }_0^{\frac{\pi }{4}}\log (1+\tan \theta )d\theta$ $=8\log 2\frac{\pi }{4}-I$ $2\mathrm{I}=2\pi \log 2$ $\mathrm{I}=\pi \log 2$