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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

thank you

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

Answered by Eulerian last updated on 13/Feb/22

(π^2 /(16))

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

Answered by qaz last updated on 13/Feb/22

∫_0 ^∞ ((xtan^(−1) x)/(1+x^4 ))dx =∫_0 ^1 ((xtan^(−1) x)/(1+x^4 ))dx+∫_1 ^∞ ((xtan^(−1) x)/(1+x^4 ))dx =∫_0 ^1 ((xtan^(−1) x)/(1+x^4 ))dx+∫_0 ^1 ((xtan^(−1) (1/x))/(1+x^4 ))dx =∫_0 ^1 ((x(tan^(−1) x+tan^(−1) (1/x)))/(1+x^4 ))dx =(π/2)∫_0 ^1 (x/(1+x^4 ))dx =(π/4)∫_0 ^1 (dx/(1+x^2 )) =(π^2 /(16))

$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{xtan}^{−\mathrm{1}} \mathrm{x}}{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{xtan}^{−\mathrm{1}} \mathrm{x}}{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }\mathrm{dx}+\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{xtan}^{−\mathrm{1}} \mathrm{x}}{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{xtan}^{−\mathrm{1}} \mathrm{x}}{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }\mathrm{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{xtan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{x}}}{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}\left(\mathrm{tan}^{−\mathrm{1}} \mathrm{x}+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{x}}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }\mathrm{dx} \\ $$$$=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}}{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }\mathrm{dx} \\ $$$$=\frac{\pi}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dx}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} } \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{16}} \\ $$

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

thank you

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

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