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Question Number 129562 by mathmax by abdo last updated on 16/Jan/21
calculate ∫_0 ^∞   (dx/((x^6 +1)^2 ))
$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{6}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$
Answered by Lordose last updated on 16/Jan/21
Ω = ∫_0 ^( ∞) (dx/((x^6 +1)^2 )) =^(x=u^(1/6) ) (1/6)∫_0 ^( ∞) (u^(−(5/6)) /((1+u)^2 ))du = (1/6)𝛃((1/6),((11)/6))     Ω = (1/6)∙((𝚪((1/6))𝚪(((11)/6)))/(𝚪(2))) = ((5π)/(18))
$$\Omega\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{6}} +\mathrm{1}\right)^{\mathrm{2}} }\:\overset{\mathrm{x}=\mathrm{u}^{\frac{\mathrm{1}}{\mathrm{6}}} } {=}\frac{\mathrm{1}}{\mathrm{6}}\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{u}^{−\frac{\mathrm{5}}{\mathrm{6}}} }{\left(\mathrm{1}+\mathrm{u}\right)^{\mathrm{2}} }\mathrm{du}\:=\:\frac{\mathrm{1}}{\mathrm{6}}\boldsymbol{\beta}\left(\frac{\mathrm{1}}{\mathrm{6}},\frac{\mathrm{11}}{\mathrm{6}}\right)\:\:\: \\ $$$$\Omega\:=\:\frac{\mathrm{1}}{\mathrm{6}}\centerdot\frac{\boldsymbol{\Gamma}\left(\frac{\mathrm{1}}{\mathrm{6}}\right)\boldsymbol{\Gamma}\left(\frac{\mathrm{11}}{\mathrm{6}}\right)}{\boldsymbol{\Gamma}\left(\mathrm{2}\right)}\:=\:\frac{\mathrm{5}\pi}{\mathrm{18}} \\ $$

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