Integration Questions

Question Number 132089 by rs4089 last updated on 11/Feb/21

$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left({x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }{dx} \\$$

Answered by Lordose last updated on 11/Feb/21

$$\Omega\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{1}}{\left(\mathrm{x}^{\mathrm{4}} −\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }\mathrm{dx}\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{3}} }{\left(\mathrm{1}+\mathrm{x}^{\mathrm{6}} \right)^{\mathrm{3}} }\mathrm{dx} \\$$$$\Omega\:\overset{\mathrm{u}=\mathrm{x}^{\mathrm{6}} } {=}\frac{\mathrm{1}}{\mathrm{6}}\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{u}^{−\frac{\mathrm{5}}{\mathrm{6}}} \left(\mathrm{1}+\mathrm{u}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)^{\mathrm{3}} }{\left(\mathrm{1}+\mathrm{u}\right)^{\mathrm{3}} }\mathrm{du}\:=\:\frac{\mathrm{1}}{\mathrm{6}}\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{2u}^{−\frac{\mathrm{5}}{\mathrm{6}}} +\mathrm{3u}^{−\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{3u}^{−\frac{\mathrm{1}}{\mathrm{6}}} }{\left(\mathrm{1}+\mathrm{u}\right)^{\mathrm{3}} }\mathrm{du} \\$$$$\Omega\:=\:\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{\beta}\left(\frac{\mathrm{1}}{\mathrm{6}},\frac{\mathrm{17}}{\mathrm{6}}\right)\:+\:\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\beta}\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{5}}{\mathrm{2}}\right)\:+\:\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\beta}\left(\frac{\mathrm{5}}{\mathrm{6}},\frac{\mathrm{13}}{\mathrm{6}}\right) \\$$$$\Omega\:=\:\frac{\mathrm{1}}{\mathrm{6}}\left(\frac{\mathrm{2}\boldsymbol{\Gamma}\left(\frac{\mathrm{1}}{\mathrm{6}}\right)\boldsymbol{\Gamma}\left(\frac{\mathrm{17}}{\mathrm{6}}\right)+\mathrm{3}\boldsymbol{\Gamma}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\boldsymbol{\Gamma}\left(\frac{\mathrm{5}}{\mathrm{2}}\right)+\mathrm{3}\boldsymbol{\Gamma}\left(\frac{\mathrm{5}}{\mathrm{6}}\right)\boldsymbol{\Gamma}\left(\frac{\mathrm{13}}{\mathrm{6}}\right)}{\boldsymbol{\Gamma}\left(\mathrm{3}\right)}\right)\: \\$$

Commented by Ar Brandon last updated on 11/Feb/21

$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{6}}\left\{\frac{\mathrm{2}\centerdot\frac{\mathrm{11}}{\mathrm{6}}\centerdot\frac{\mathrm{5}}{\mathrm{6}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{6}}\right)\Gamma\left(\frac{\mathrm{5}}{\mathrm{6}}\right)+\mathrm{3}\centerdot\frac{\mathrm{3}}{\mathrm{2}}\centerdot\frac{\mathrm{1}}{\mathrm{2}}\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{3}\centerdot\frac{\mathrm{7}}{\mathrm{6}}\centerdot\frac{\mathrm{1}}{\mathrm{6}}\Gamma\left(\frac{\mathrm{5}}{\mathrm{6}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{6}}\right)}{\mathrm{2}\centerdot\mathrm{1}}\right\} \\$$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{12}}\left\{\frac{\mathrm{55}\pi}{\mathrm{9}}+\frac{\mathrm{9}\pi}{\mathrm{4}}+\frac{\mathrm{7}\pi}{\mathrm{6}}\right\}=\frac{\mathrm{1}}{\mathrm{12}}\centerdot\frac{\mathrm{220}+\mathrm{81}+\mathrm{42}}{\mathrm{36}}\pi=\frac{\mathrm{343}}{\mathrm{432}}\pi \\$$

Commented by Lordose last updated on 11/Feb/21

$$\mathrm{Finishing}\:\mathrm{touches},\:\mathrm{hahaha} \\$$

Commented by Ar Brandon last updated on 11/Feb/21

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