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