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


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Question Number 142869 by mathmax by abdo last updated on 06/Jun/21

$calculate \int_{- \infty}^{+ \infty} \frac{x^{2} dx}{{(x^{2} - x + 3)}^{2}}$
	Answered by Ar Brandon last updated on 06/Jun/21

	$Ψ = \int_{- \infty}^{\infty} \frac{x^{2} dx}{{(x^{2} - x + 3)}^{2}}$ $\int \frac{x^{2} dx}{{(x^{2} - x + 3)}^{2}} = \frac{ax + b}{x^{2} - x + 3} + \int \frac{cx + d}{x^{2} - x + 3} dx$ $\Rightarrow \frac{x^{2} dx}{{(x^{2} - x + 3)}^{2}} = \frac{a (x^{2} - x + 3) - (ax + b) (2 x - 1)}{{(x^{2} - x + 3)}^{2}} + \frac{(cx + d) (x^{2} - x + 3)}{{(x^{2} - x + 3)}^{2}}$ $c = 0, a - 2 a - c + d = 1 \Rightarrow d = a + 1, - a + a - 2 b + 3 c - d = 0 \Rightarrow d = - 2 b$ $3 a + b + 3 d = 0 \Rightarrow 6 a - a - 1 + 6 a + 6 = 0 \Rightarrow a = - \frac{5}{11}, d = \frac{6}{11}, b = - \frac{3}{11}$ $Ψ = {[- \frac{5 x + 3}{11 (x^{2} - x + 3)} + \frac{6}{11} \int \frac{dx}{x^{2} - x + 3}]}_{- \infty}^{\infty}$ $= \frac{6}{11} \int_{- \infty}^{\infty} \frac{dx}{{(x - \frac{1}{2})}^{2} + \frac{11}{4}} = {[\frac{6}{11} \cdot \frac{2}{\sqrt{11}} \arctan (\frac{2 x - 1}{\sqrt{11}})]}_{- \infty}^{\infty} = \frac{12 \sqrt{11}}{121} π$
	Commented by mathmax by abdo last updated on 08/Jun/21

	$thanks sir your answer is cool!$
	Answered by mathmax by abdo last updated on 08/Jun/21
	$Φ=∫_(−∞) ^(+∞) ((x^2 dx)/((x^2 −x+3)^2 )) let ϕ(z)=(z^2 /((z^2 −z+3)^2 )) poles of ϕ? z^2 −z+3=0 →Δ=1−12=−11 →z_1 =((1+i(√(11)))/2) and z_2 =((1−i(√(11)))/2) ϕ(z)=(z^2 /((z−z_1 )^2 (z−z_2 )^2 )) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπRes(ϕ,z_1 ) Res(ϕ,z_1 )=lim_(z→z_1 ) (1/((2−1)!)){(z−z_1 )^2 ϕ(z)}^((1)) =lim_(z→z_1 ) {(z^2 /((z−z_2 )^2 ))}^((1)) =lim_(z→z_1 ) ((2z(z−z_2 )^2 −2z^2 (z−z_2 ))/((z−z_2 )^4 )) =lim_(z→z_1 ) ((2z(z−z_2 )−2z^2 )/((z−z_2 )^3 )) =((2z_1 (z_1 −z_2 )−2z_1 ^2 )/((z_1 −z_2 )^3 )) =((−2z_1 z_2 )/((z_1 −z_2 )^3 )) =((−6)/((i(√(11)))^3 )) =((−6)/(−i11(√(11))))=((−6i)/(11(√(11)))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz=2iπ×((−6i)/(11(√(11)))) =((12π)/(11(√(11)))) ⇒ Φ=((12π)/(11(√(11))))$
	$Φ = \int_{- \infty}^{+ \infty} \frac{x^{2} dx}{{(x^{2} - x + 3)}^{2}} let φ (z) = \frac{z^{2}}{{(z^{2} - z + 3)}^{2}} poles of φ ?$ $z^{2} - z + 3 = 0 \to Δ = 1 - 12 = - 11 \to z_{1} = \frac{1 + i \sqrt{11}}{2} and z_{2} = \frac{1 - i \sqrt{11}}{2}$ $φ (z) = \frac{z^{2}}{{(z - z_{1})}^{2} {(z - z_{2})}^{2}} residus theorem give$ $\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π Res (φ, z_{1})$ $Res (φ, z_{1}) = \lim_{z \to z_{1}} \frac{1}{(2 - 1)!} {{(z - z_{1})}^{2} φ (z)}^{(1)}$ $= \lim_{z \to z_{1}} {\frac{z^{2}}{{(z - z_{2})}^{2}}}^{(1)} = \lim_{z \to z_{1}} \frac{2 z {(z - z_{2})}^{2} - {2 z}^{2} (z - z_{2})}{{(z - z_{2})}^{4}}$ $= \lim_{z \to z_{1}} \frac{2 z (z - z_{2}) - {2 z}^{2}}{{(z - z_{2})}^{3}} = \frac{{2 z}_{1} (z_{1} - z_{2}) - {2 z}_{1}^{2}}{{(z_{1} - z_{2})}^{3}}$ $= \frac{- {2 z}_{1} z_{2}}{{(z_{1} - z_{2})}^{3}} = \frac{- 6}{{(i \sqrt{11})}^{3}} = \frac{- 6}{- i 11 \sqrt{11}} = \frac{- 6 i}{11 \sqrt{11}} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π \times \frac{- 6 i}{11 \sqrt{11}} = \frac{12 π}{11 \sqrt{11}} \Rightarrow$ $Φ = \frac{12 π}{11 \sqrt{11}}$
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