Question Number 166093 by peter frank last updated on 13/Feb/22
Commented by Eulerian last updated on 13/Feb/22
$$ \\ $$
Commented by Eulerian last updated on 13/Feb/22
Commented by peter frank last updated on 13/Feb/22
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by Eulerian last updated on 13/Feb/22
$$\:\frac{\pi^{\mathrm{2}} }{\mathrm{16}} \\ $$
Answered by qaz last updated on 13/Feb/22
$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{xtan}^{−\mathrm{1}} \mathrm{x}}{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{xtan}^{−\mathrm{1}} \mathrm{x}}{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }\mathrm{dx}+\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{xtan}^{−\mathrm{1}} \mathrm{x}}{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{xtan}^{−\mathrm{1}} \mathrm{x}}{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }\mathrm{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{xtan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{x}}}{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}\left(\mathrm{tan}^{−\mathrm{1}} \mathrm{x}+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{x}}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }\mathrm{dx} \\ $$$$=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}}{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }\mathrm{dx} \\ $$$$=\frac{\pi}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dx}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} } \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{16}} \\ $$
Commented by peter frank last updated on 13/Feb/22
$$\mathrm{thank}\:\mathrm{you} \\ $$