Question Number 153114 by peter frank last updated on 04/Sep/21
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{log}\:\left(\mathrm{1}+\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$
Answered by puissant last updated on 04/Sep/21
$${x}={tan}\left({u}\right)\:\rightarrow\:{dx}=\mathrm{1}+{tan}^{\mathrm{2}} {udu} \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{ln}\left(\mathrm{1}+{tan}\left({u}\right)\right)}{\mathrm{1}+{tan}^{\mathrm{2}} {u}}\left(\mathrm{1}+{tan}^{\mathrm{2}} {u}\right){du} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\mathrm{1}+{tan}\left({u}\right)\right){du} \\ $$$${u}=\frac{\pi}{\mathrm{4}}−{t}\:\rightarrow\:{du}=−{dt} \\ $$$${I}=\int_{\frac{\pi}{\mathrm{4}}} ^{\mathrm{0}} {ln}\left(\mathrm{1}+{tan}\left(\frac{\pi}{\mathrm{4}}−{t}\right)\right)\left(−{dt}\right) \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\mathrm{1}+\frac{\mathrm{1}−{tan}\left({t}\right)}{\mathrm{1}+{tan}\left({t}\right)}\right){dt} \\ $$$$\Rightarrow\:{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\mathrm{2}{dt}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\mathrm{1}+{tan}\left({t}\right)\right){dt} \\ $$$$\Rightarrow\:\mathrm{2}{I}=\frac{\pi}{\mathrm{4}}{ln}\mathrm{2} \\ $$$$ \\ $$$$\therefore\because\:\:{I}=\frac{\pi}{\mathrm{8}}{ln}\mathrm{2}.. \\ $$
Commented by peter frank last updated on 04/Sep/21
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by Ar Brandon last updated on 04/Sep/21
$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx},\:{x}=\mathrm{tan}\vartheta \\ $$$$\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{1}+\mathrm{tan}\vartheta\right)\mathrm{d}\vartheta \\ $$$$\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{cos}\vartheta+\mathrm{sin}\vartheta\right){d}\vartheta−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{cos}\vartheta\right){d}\vartheta \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left({G}−\frac{\pi\mathrm{ln2}}{\mathrm{4}}\right)−\left(\frac{{G}}{\mathrm{2}}−\frac{\pi\mathrm{ln2}}{\mathrm{4}}\right)=\frac{\pi\mathrm{ln2}}{\mathrm{8}} \\ $$
Commented by peter frank last updated on 05/Sep/21
$$\mathrm{thank}\:\mathrm{you} \\ $$