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Question Number 192280 by Mingma last updated on 14/May/23

Answered by witcher3 last updated on 14/May/23

$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{({\mathrm{x}}^4-{\mathrm{x}}^2+1)}^3}$$$$={\int }_0^{\mathrm{\infty }}\frac{{({\mathrm{x}}^2+1)}^3}{{({\mathrm{x}}^2+1)}^3{({\mathrm{x}}^4-{\mathrm{x}}^2+1)}^3}\mathrm{dx}$$$$={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^6+{3\mathrm{x}}^4+{3\mathrm{x}}^2+1}{{({\mathrm{x}}^6+1)}^3}\mathrm{dx}$$$${\mathrm{x}}^6=\mathrm{u}$$$$={\int }_0^{\mathrm{\infty }}\frac{\mathrm{u}+{3\mathrm{u}}^{\frac{2}{3}}+{3\mathrm{u}}^{\frac{1}{3}}+1}{{(\mathrm{u}+1)}^3}.\frac{{\mathrm{u}}^{-\frac{5}{6}}}{6}\mathrm{du}$$$$=\frac{1}{6}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{u}}^{\frac{1}{6}}}{{(1+\mathrm{u})}^3}+\frac{{3\mathrm{u}}^{\frac{-1}{6}}}{{(1+\mathrm{u})}^3}+\frac{{3\mathrm{u}}^{-\frac{3}{6}}}{{(1+\mathrm{u})}^3}+\frac{{\mathrm{u}}^{-\frac{5}{6}}}{{(1+\mathrm{u})}^3}\mathrm{du}$$$$\beta (\mathrm{x},\mathrm{y})={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\mathrm{x}-1}}{{(1+\mathrm{t})}^{\mathrm{x}+\mathrm{y}}}$$$$\mathrm{\Omega }=\frac{\beta (\frac{7}{6},\frac{11}{6})+3\beta (\frac{5}{6},\frac{13}{6})+3\beta (\frac{3}{6},\frac{15}{6})+\beta (\frac{1}{6},\frac{17}{6})}{6}$$$$=\frac{\mathrm{\Gamma }\left(\frac{7}{6}\right)\mathrm{\Gamma }\left(\frac{11}{6}\right)+3\mathrm{\Gamma }\left(\frac{5}{6}\right)\mathrm{\Gamma }\left(\frac{13}{6}\right)+3\mathrm{\Gamma }\left(\frac{3}{6}\right)\mathrm{\Gamma }\left(\frac{15}{6}\right)+\mathrm{\Gamma }\left(\frac{1}{6}\right)\mathrm{\Gamma }\left(\frac{17}{6}\right)}{12}$$$$\mathrm{\Gamma }(3-\mathrm{x})=(2-\mathrm{x})(1-\mathrm{x})\mathrm{\Gamma }(1-\mathrm{x})$$$$\mathrm{\Gamma }\left(\mathrm{x}\right)\mathrm{\Gamma }(1-\mathrm{x})=\frac{\pi }{\sin \left(\pi \mathrm{x}\right)}$$$$\mathrm{\Omega }=\frac{\pi }{12\sin \left(\frac{\pi }{6}\right)}.\frac{1}{6}.\frac{5}{6}+\frac{3\pi }{12\sin \left(\frac{5\pi }{6}\right)}.\frac{7}{6}.\frac{1}{6}+\frac{3\pi }{\sin \left(\frac{\pi }{6}\right)}.\frac{11}{6}.\frac{5}{6}$$$$+\frac{\pi }{12\sin \left(\frac{\pi }{6}\right)}.\frac{11}{6}.\frac{5}{6}$$$$$$

Commented by Mingma last updated on 14/May/23

Nice!

Answered by MJS_new last updated on 14/May/23

$$\int \frac{dx}{{(x^4-x^2+1)}^3}=$$$$\left[{\mathrm{Ostrogradski}}^{\prime }\mathrm{s}\mathrm{Method}\right]$$$$=\frac{x(7x^6-5x^4+7x^2+4)}{{(x^4-x^2+1)}^2}+\frac{1}{24}\int \frac{7x^2+20}{x^4-x^2+1}dx=$$$$[\mathrm{decomposing}\mathrm{etc}.]$$$$=\frac{x(7x^6-5x^4+7x^2+4)}{{(x^4-x^2+1)}^2}+\frac{13\sqrt{3}}{288}\ln \frac{x^2+\sqrt{3}x+1}{x^2-\sqrt{3}x+1}+\frac{9}{16}(\arctan (2x+\sqrt{3})+\arctan (2x+\sqrt{3}))+C$$$$\Rightarrow \mathrm{answer}\mathrm{is}\frac{9\pi }{16}$$

Commented by Mingma last updated on 14/May/23

Exactly!

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