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Question Number 129562 by mathmax by abdo last updated on 16/Jan/21

$$\mathrm{calculate}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{({\mathrm{x}}^6+1)}^2}$$

Answered by Lordose last updated on 16/Jan/21

$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{({\mathrm{x}}^6+1)}^2}\stackrel{\mathrm{x}={\mathrm{u}}^{\frac{1}{6}}}{=}\frac{1}{6}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{u}}^{-\frac{5}{6}}}{{(1+\mathrm{u})}^2}\mathrm{du}=\frac{1}{6}\mathit{\beta }(\frac{1}{6},\frac{11}{6})$$$$\mathrm{\Omega }=\frac{1}{6}\cdot \frac{\mathbf{\Gamma }\left(\frac{1}{6}\right)\mathbf{\Gamma }\left(\frac{11}{6}\right)}{\mathbf{\Gamma }\left(2\right)}=\frac{5\pi }{18}$$

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