Question Number 137770 by peter frank last updated on 06/Apr/21
Answered by Ñï= last updated on 06/Apr/21
$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} {da}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}}{\left(\mathrm{1}+{ax}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{a}^{\mathrm{2}} }{da}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{−{a}}{\mathrm{1}+{ax}}+\frac{{x}+{a}}{\mathrm{1}+{x}^{\mathrm{2}} }\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{a}^{\mathrm{2}} }{da}\left(−{ln}\left(\mathrm{1}+{ax}\right)+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)+{a}\mathrm{tan}^{−\mathrm{1}} {x}\right)_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{a}^{\mathrm{2}} }\left(−{ln}\left(\mathrm{1}+{a}\right)+\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{2}+\frac{\pi}{\mathrm{4}}{a}\right){da} \\ $$$$=\frac{\pi}{\mathrm{4}}{ln}\mathrm{2}−{I} \\ $$$$\Rightarrow{I}=\frac{\pi}{\mathrm{8}}{ln}\mathrm{2} \\ $$