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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

$$\frac{{\pi }^2}{16}$$

Answered by qaz last updated on 13/Feb/22

$${\int }_0^{\mathrm{\infty }}\frac{{\mathrm{xtan}}^{-1}\mathrm{x}}{1+{\mathrm{x}}^4}\mathrm{dx}$$$$={\int }_0^1\frac{{\mathrm{xtan}}^{-1}\mathrm{x}}{1+{\mathrm{x}}^4}\mathrm{dx}+{\int }_1^{\mathrm{\infty }}\frac{{\mathrm{xtan}}^{-1}\mathrm{x}}{1+{\mathrm{x}}^4}\mathrm{dx}$$$$={\int }_0^1\frac{{\mathrm{xtan}}^{-1}\mathrm{x}}{1+{\mathrm{x}}^4}\mathrm{dx}+{\int }_0^1\frac{{\mathrm{xtan}}^{-1}\frac{1}{\mathrm{x}}}{1+{\mathrm{x}}^4}\mathrm{dx}$$$$={\int }_0^1\frac{\mathrm{x}({\tan }^{-1}\mathrm{x}+{\tan }^{-1}\frac{1}{\mathrm{x}})}{1+{\mathrm{x}}^4}\mathrm{dx}$$$$=\frac{\pi }{2}{\int }_0^1\frac{\mathrm{x}}{1+{\mathrm{x}}^4}\mathrm{dx}$$$$=\frac{\pi }{4}{\int }_0^1\frac{\mathrm{dx}}{1+{\mathrm{x}}^2}$$$$=\frac{{\pi }^2}{16}$$

Commented by peter frank last updated on 13/Feb/22

$$\mathrm{thank}\mathrm{you}$$

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