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Question Number 192280 by Mingma last updated on 14/May/23
Answered by witcher3 last updated on 14/May/23
Ω=∫_0 ^∞ (dx/((x^4 −x^2 +1)^3 )) =∫_0 ^∞ (((x^2 +1)^3 )/((x^2 +1)^3 (x^4 −x^2 +1)^3 ))dx =∫_0 ^∞ ((x^6 +3x^4 +3x^2 +1)/((x^6 +1)^3 ))dx x^6 =u =∫_0 ^∞ ((u+3u^(2/3) +3u^(1/3) +1)/((u+1)^3 )).(u^(−(5/6)) /6)du =(1/6)∫_0 ^∞ (u^(1/6) /((1+u)^3 ))+((3u^((−1)/6) )/((1+u)^3 ))+((3u^(−(3/6)) )/((1+u)^3 ))+(u^(−(5/6)) /((1+u)^3 ))du β(x,y)=∫_0 ^∞ (t^(x−1) /((1+t)^(x+y) )) Ω=((β((7/6),((11)/6))+3β((5/6),((13)/6))+3β((3/6),((15)/6))+β((1/6),((17)/6)))/6) =((Γ((7/6))Γ(((11)/6))+3Γ((5/6))Γ(((13)/6))+3Γ((3/6))Γ(((15)/6))+Γ((1/6))Γ(((17)/6)))/(12)) Γ(3−x)=(2−x)(1−x)Γ(1−x) Γ(x)Γ(1−x)=(π/(sin(πx))) Ω=(π/(12sin((π/6)))).(1/6).(5/6)+((3π)/(12sin(((5π)/6)))).(7/6).(1/6)+((3π)/(sin((π/6)))).((11)/6).(5/6) +(π/(12sin((π/6)))).((11)/6).(5/6)
$$\Omega=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{4}} −\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} \left(\mathrm{x}^{\mathrm{4}} −\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{x}^{\mathrm{6}} +\mathrm{3x}^{\mathrm{4}} +\mathrm{3x}^{\mathrm{2}} +\mathrm{1}}{\left(\mathrm{x}^{\mathrm{6}} +\mathrm{1}\right)^{\mathrm{3}} }\mathrm{dx} \\ $$$$\mathrm{x}^{\mathrm{6}} =\mathrm{u} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{u}+\mathrm{3u}^{\frac{\mathrm{2}}{\mathrm{3}}} +\mathrm{3u}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{1}}{\left(\mathrm{u}+\mathrm{1}\right)^{\mathrm{3}} }.\frac{\mathrm{u}^{−\frac{\mathrm{5}}{\mathrm{6}}} }{\mathrm{6}}\mathrm{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{u}^{\frac{\mathrm{1}}{\mathrm{6}}} }{\left(\mathrm{1}+\mathrm{u}\right)^{\mathrm{3}} }+\frac{\mathrm{3u}^{\frac{−\mathrm{1}}{\mathrm{6}}} }{\left(\mathrm{1}+\mathrm{u}\right)^{\mathrm{3}} }+\frac{\mathrm{3u}^{−\frac{\mathrm{3}}{\mathrm{6}}} }{\left(\mathrm{1}+\mathrm{u}\right)^{\mathrm{3}} }+\frac{\mathrm{u}^{−\frac{\mathrm{5}}{\mathrm{6}}} }{\left(\mathrm{1}+\mathrm{u}\right)^{\mathrm{3}} }\mathrm{du} \\ $$$$\beta\left(\mathrm{x},\mathrm{y}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}^{\mathrm{x}−\mathrm{1}} }{\left(\mathrm{1}+\mathrm{t}\right)^{\mathrm{x}+\mathrm{y}} } \\ $$$$\Omega=\frac{\beta\left(\frac{\mathrm{7}}{\mathrm{6}},\frac{\mathrm{11}}{\mathrm{6}}\right)+\mathrm{3}\beta\left(\frac{\mathrm{5}}{\mathrm{6}},\frac{\mathrm{13}}{\mathrm{6}}\right)+\mathrm{3}\beta\left(\frac{\mathrm{3}}{\mathrm{6}},\frac{\mathrm{15}}{\mathrm{6}}\right)+\beta\left(\frac{\mathrm{1}}{\mathrm{6}},\frac{\mathrm{17}}{\mathrm{6}}\right)}{\mathrm{6}} \\ $$$$=\frac{\Gamma\left(\frac{\mathrm{7}}{\mathrm{6}}\right)\Gamma\left(\frac{\mathrm{11}}{\mathrm{6}}\right)+\mathrm{3}\Gamma\left(\frac{\mathrm{5}}{\mathrm{6}}\right)\Gamma\left(\frac{\mathrm{13}}{\mathrm{6}}\right)+\mathrm{3}\Gamma\left(\frac{\mathrm{3}}{\mathrm{6}}\right)\Gamma\left(\frac{\mathrm{15}}{\mathrm{6}}\right)+\Gamma\left(\frac{\mathrm{1}}{\mathrm{6}}\right)\Gamma\left(\frac{\mathrm{17}}{\mathrm{6}}\right)}{\mathrm{12}} \\ $$$$\Gamma\left(\mathrm{3}−\mathrm{x}\right)=\left(\mathrm{2}−\mathrm{x}\right)\left(\mathrm{1}−\mathrm{x}\right)\Gamma\left(\mathrm{1}−\mathrm{x}\right) \\ $$$$\Gamma\left(\mathrm{x}\right)\Gamma\left(\mathrm{1}−\mathrm{x}\right)=\frac{\pi}{\mathrm{sin}\left(\pi\mathrm{x}\right)} \\ $$$$\Omega=\frac{\pi}{\mathrm{12sin}\left(\frac{\pi}{\mathrm{6}}\right)}.\frac{\mathrm{1}}{\mathrm{6}}.\frac{\mathrm{5}}{\mathrm{6}}+\frac{\mathrm{3}\pi}{\mathrm{12sin}\left(\frac{\mathrm{5}\pi}{\mathrm{6}}\right)}.\frac{\mathrm{7}}{\mathrm{6}}.\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{3}\pi}{\mathrm{sin}\left(\frac{\pi}{\mathrm{6}}\right)}.\frac{\mathrm{11}}{\mathrm{6}}.\frac{\mathrm{5}}{\mathrm{6}} \\ $$$$+\frac{\pi}{\mathrm{12sin}\left(\frac{\pi}{\mathrm{6}}\right)}.\frac{\mathrm{11}}{\mathrm{6}}.\frac{\mathrm{5}}{\mathrm{6}} \\ $$$$ \\ $$
Commented by Mingma last updated on 14/May/23
Nice!
Answered by MJS_new last updated on 14/May/23
∫(dx/((x^4 −x^2 +1)^3 ))= [Ostrogradski′s Method] =((x(7x^6 −5x^4 +7x^2 +4))/((x^4 −x^2 +1)^2 ))+(1/(24))∫((7x^2 +20)/(x^4 −x^2 +1))dx= [decomposing etc.] =((x(7x^6 −5x^4 +7x^2 +4))/((x^4 −x^2 +1)^2 ))+((13(√3))/(288))ln ((x^2 +(√3)x+1)/(x^2 −(√3)x+1)) +(9/(16))(arctan (2x+(√3)) +arctan (2x+(√3))) +C ⇒ answer is ((9π)/(16))
$$\int\frac{{dx}}{\left({x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }= \\ $$$$\:\:\:\:\:\left[\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{Method}\right] \\ $$$$=\frac{{x}\left(\mathrm{7}{x}^{\mathrm{6}} −\mathrm{5}{x}^{\mathrm{4}} +\mathrm{7}{x}^{\mathrm{2}} +\mathrm{4}\right)}{\left({x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{24}}\int\frac{\mathrm{7}{x}^{\mathrm{2}} +\mathrm{20}}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}}{dx}= \\ $$$$\:\:\:\:\:\left[\mathrm{decomposing}\:\mathrm{etc}.\right] \\ $$$$=\frac{{x}\left(\mathrm{7}{x}^{\mathrm{6}} −\mathrm{5}{x}^{\mathrm{4}} +\mathrm{7}{x}^{\mathrm{2}} +\mathrm{4}\right)}{\left({x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{13}\sqrt{\mathrm{3}}}{\mathrm{288}}\mathrm{ln}\:\frac{{x}^{\mathrm{2}} +\sqrt{\mathrm{3}}{x}+\mathrm{1}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{3}}{x}+\mathrm{1}}\:+\frac{\mathrm{9}}{\mathrm{16}}\left(\mathrm{arctan}\:\left(\mathrm{2}{x}+\sqrt{\mathrm{3}}\right)\:+\mathrm{arctan}\:\left(\mathrm{2}{x}+\sqrt{\mathrm{3}}\right)\right)\:+{C} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\frac{\mathrm{9}\pi}{\mathrm{16}} \\ $$
Commented by Mingma last updated on 14/May/23
Exactly!

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