calculate ∫_(−∞) ^∞ ((x^2 −3)/((x^2 −x+1)^3 ))dx


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Question Number 96956 by mathmax by abdo last updated on 05/Jun/20

$calculate \int_{- \infty}^{\infty} \frac{x^{2} - 3}{{(x^{2} - x + 1)}^{3}} dx$
	Answered by MJS last updated on 06/Jun/20

	$Ostrogradski leads to$ $\int \frac{x^{2} - 3}{{(x^{2} - x + 1)}^{3}} d x =$ $= - \frac{10 x^{3} - 15 x^{2} + 22 x - 7}{6 {(x^{2} - x + 1)}^{2}} - \frac{5}{3} \int \frac{d x}{x^{2} - x + 1} =$ $= - \frac{10 x^{3} - 15 x^{2} + 22 x - 7}{6 {(x^{2} - x + 1)}^{2}} - \frac{10 \sqrt{3}}{9} \arctan \frac{\sqrt{3} (2 x - 1)}{3} + C$ $\Rightarrow$ $\underset{- \infty}{\int^{+ \infty}} \frac{x^{2} - 3}{{(x^{2} - x + 1)}^{3}} d x = - \frac{10 \sqrt{3}}{9} π$
	Commented by abdomathmax last updated on 06/Jun/20

	$thankx sir mjs$
	Answered by abdomathmax last updated on 06/Jun/20
	$A =∫_(−∞) ^(+∞) ((x^2 −3)/((x^2 −x+1)^3 ))dx let ϕ(z) =((z^2 −3)/((x^2 −x+1)^3 )) x^2 −x+1 =0 →Δ =−3 ⇒x_1 =((1+i(√3))/2) =e^((iπ)/3) x_2 =((1−i(√3))/2) =e^(−((iπ)/3)) ⇒ϕ(z) =((z^2 −3)/((z−e^((iπ)/3) )^3 (z−e^(−((iπ)/3)) )^3 )) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,e^((iπ)/3) ) Res(ϕ,e^((iπ)/3) ) =lim_(z→e^((iπ)/3) ) (1/((3−1)!)){(z−e^((iπ)/3) )^3 ϕ(z)}^((2)) =lim_(z→e^((iπ)/3) ) (1/2){((z^2 −3)/((z−e^(−((iπ)/3)) )^3 ))}^((2)) =(1/2)lim_(z→e^((iπ)/3) ) { ((2z(z−e^(−((iπ)/3)) )^3 −3(z−e^(−((iπ)/3)) )^2 (z^2 −3))/((z−e^(−((iπ)/3)) )^6 ))}^((1)) =(1/2)lim_(z→e^((iπ)/3) ) {((2z(z−e^(−((iπ)/3)) )−3(z^2 −3))/((z−e^(−((iπ)/3)) )^4 ))}^((1)) ....be continued....$
	$A = \int_{- \infty}^{+ \infty} \frac{x^{2} - 3}{{(x^{2} - x + 1)}^{3}} dx let φ (z) = \frac{z^{2} - 3}{{(x^{2} - x + 1)}^{3}}$ $x^{2} - x + 1 = 0 \to Δ = - 3 \Rightarrow x_{1} = \frac{1 + i \sqrt{3}}{2} = e^{\frac{i π}{3}}$ $x_{2} = \frac{1 - i \sqrt{3}}{2} = e^{- \frac{i π}{3}} \Rightarrow φ (z) = \frac{z^{2} - 3}{{(z - e^{\frac{i π}{3}})}^{3} {(z - e^{- \frac{i π}{3}})}^{3}}$ $\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π Res (φ, e^{\frac{i π}{3}})$ $Res (φ, e^{\frac{i π}{3}}) = \lim_{z \to e^{\frac{i π}{3}}} \frac{1}{(3 - 1)!} {{(z - e^{\frac{i π}{3}})}^{3} φ (z)}^{(2)}$ $= \lim_{z \to e^{\frac{i π}{3}}} \frac{1}{2} {\frac{z^{2} - 3}{{(z - e^{- \frac{i π}{3}})}^{3}}}^{(2)}$ $= \frac{1}{2} \lim_{z \to e^{\frac{i π}{3}}} {\frac{2 z {(z - e^{- \frac{i π}{3}})}^{3} - 3 {(z - e^{- \frac{i π}{3}})}^{2} (z^{2} - 3)}{{(z - e^{- \frac{i π}{3}})}^{6}}}^{(1)}$ $= \frac{1}{2} \lim_{z \to e^{\frac{i π}{3}}} {\frac{2 z (z - e^{- \frac{i π}{3}}) - 3 (z^{2} - 3)}{{(z - e^{- \frac{i π}{3}})}^{4}}}^{(1)}$ $. . . . be continued . . . .$
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