Mathematics - Definite Integration Question with Solution | TestHub

Let $x=\tan \theta$. Then $dx={\sec }^2\theta d\theta$. When $x=0,\theta =0$. When $x=1,\theta =\pi /4$. The integral becomes $I={\int }_0^{\pi /4}\frac{\ln (1+\tan \theta )}{1+{\tan }^2\theta }{\sec }^2\theta d\theta ={\int }_0^{\pi /4}\ln (1+\tan \theta )d\theta$. Using the property ${\int }_0^{a}g(x)dx={\int }_0^{a}g(a-x)dx$: $I={\int }_0^{\pi /4}\ln (1+\tan (\pi /4-\theta ))d\theta$. Since $\tan (\pi /4-\theta )=\frac{1-\tan \theta }{1+\tan \theta }$, we have $1+\tan (\pi /4-\theta )=1+\frac{1-\tan \theta }{1+\tan \theta }=\frac{2}{1+\tan \theta }$. So, $I={\int }_0^{\pi /4}\ln \left(\frac{2}{1+\tan \theta }\right)d\theta ={\int }_0^{\pi /4}(\ln 2-\ln (1+\tan \theta ))d\theta$.$I=\ln 2{\int }_0^{\pi /4}d\theta -{\int }_0^{\pi /4}\ln (1+\tan \theta )d\theta =\ln 2[\theta {]}_0^{\pi /4}-I$. Thus, $I=\frac{\pi }{4}\ln 2-I$, which gives $2I=\frac{\pi }{4}\ln 2$, so $I=\frac{\pi }{8}\ln 2$ (approximately 0.272198).