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


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Question Number 115449 by mathmax by abdo last updated on 25/Sep/20

$calculate \int_{- \infty}^{\infty} \frac{x^{2}}{{(x^{2} - x + 1)}^{3}} dx$
	Answered by MJS_new last updated on 25/Sep/20

	$\int \frac{x^{2}}{{(x^{2} - x + 1)}^{3}} d x =$ $[Ostrogeadski]$ $= \frac{2 x^{3} - 3 x^{2} + 2 x - 2}{6 {(x^{2} - x + 1)}^{2}} + \frac{1}{3} \int \frac{d x}{x^{2} - x + 1} =$ $= \frac{2 x^{3} - 3 x^{2} + 2 x - 2}{6 {(x^{2} - x + 1)}^{2}} + \frac{2 \sqrt{3}}{9} \arctan \frac{\sqrt{3} (2 x - 1)}{3} + C$ $\Rightarrow answer is \frac{2 \sqrt{3}}{9} π$
	Commented by mathmax by abdo last updated on 26/Sep/20

	$thank you sir$
	Answered by mathmax by abdo last updated on 26/Sep/20
	$A =∫_(−∞) ^(+∞) (x^2 /((x^2 −x+1)^3 ))dx we have x^2 −x+1 =x^2 −2(x/2) +(1/4) +(3/4) =(x−(1/2))^2 +(3/4) we do the changement x−(1/2) =((√3)/2)t ⇒ A =∫_(−∞) ^(+∞) ((((1/2)+((√3)/2)t)^2 )/(((3/4)(1+t^2 ))^3 )).((√3)/2)dt =((√3)/8).(4^3 /3^3 ) ∫_(−∞) ^(+∞) ((((√3)t+1)^2 )/((t^2 +1)^3 ))dt =((8(√3))/(27)) ∫_(−∞) ^(+∞) ((3t^2 +2(√3)t +1)/((t^2 +1)^3 ))dt =((8(√3))/(27))∫_(−∞) ^(+∞) ((3t^2 +1)/((t^2 +1)^3 ))dt +((8(√3))/(27))∫_(−∞) ^(+∞) ((2(√3)t)/((t^2 +1)^3 ))dt(→=0) =((8(√3))/(27)) ∫_(−∞) ^(+∞) ((3t^2 +1)/((t^2 +1)^3 ))dt let ϕ(z) =((3z^2 +1)/((z^2 +1)^3 )) ⇒ϕ(z) =((3z^2 +1)/((z−i)^3 (z+i)^3 )) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) and Res(ϕ,i) =lim_(z→i) (1/((3−1)!)){(z−i)^3 ϕ(z)}^((2)) =(1/2)lim_(z→i) {((3z^2 +1)/((z+i)^3 ))}^((2)) =(1/2)lim_(z→i) {((6z(z+i)^3 −3(z+i)^2 (3z^2 +1))/((z+i)^6 ))}^((1)) =(1/2)lim_(z→i) {((6z(z+i)−3(3z^2 +1))/((z+i)^4 ))}^((1)) =(1/2)lim_(z→i) {((−3z^2 +6iz−3)/((z+i)^4 ))}^((1)) =(1/2)lim_(z→i) (((−6z+6i)(z+i)^4 −4(z+i)^3 (−3z^2 +6iz −3))/((z+i)^8 )) =(1/2)lim_(z→i) (((−6z+6i)(z+i)−4(−3z^2 +6iz −3))/((z+i)^5 )) =(1/2).((−4(3−9))/((2i)^5 )) =((24)/(2.2^5 .i)) =((12)/(2^5 i)) =((2^2 .3)/(2^5 i)) =(3/(8i)) ⇒∫_(−∞) ^(+∞) ϕ(z)dz =2iπ.(3/(8i)) =((3π)/4) ⇒A =((8(√3))/(27)).((3π)/4) =((2π(√3))/9)$
	$A = \int_{- \infty}^{+ \infty} \frac{x^{2}}{{(x^{2} - x + 1)}^{3}} dx we have x^{2} - x + 1 = x^{2} - 2 \frac{x}{2} + \frac{1}{4} + \frac{3}{4}$ $= {(x - \frac{1}{2})}^{2} + \frac{3}{4} we do the changement x - \frac{1}{2} = \frac{\sqrt{3}}{2} t \Rightarrow$ $A = \int_{- \infty}^{+ \infty} \frac{{(\frac{1}{2} + \frac{\sqrt{3}}{2} t)}^{2}}{{(\frac{3}{4} (1 + t^{2}))}^{3}} . \frac{\sqrt{3}}{2} dt = \frac{\sqrt{3}}{8} . \frac{4^{3}}{3^{3}} \int_{- \infty}^{+ \infty} \frac{{(\sqrt{3} t + 1)}^{2}}{{(t^{2} + 1)}^{3}} dt$ $= \frac{8 \sqrt{3}}{27} \int_{- \infty}^{+ \infty} \frac{{3 t}^{2} + 2 \sqrt{3} t + 1}{{(t^{2} + 1)}^{3}} dt = \frac{8 \sqrt{3}}{27} \int_{- \infty}^{+ \infty} \frac{{3 t}^{2} + 1}{{(t^{2} + 1)}^{3}} dt + \frac{8 \sqrt{3}}{27} \int_{- \infty}^{+ \infty} \frac{2 \sqrt{3} t}{{(t^{2} + 1)}^{3}} dt (\to= 0)$ $= \frac{8 \sqrt{3}}{27} \int_{- \infty}^{+ \infty} \frac{{3 t}^{2} + 1}{{(t^{2} + 1)}^{3}} dt let φ (z) = \frac{{3 z}^{2} + 1}{{(z^{2} + 1)}^{3}} \Rightarrow φ (z) = \frac{{3 z}^{2} + 1}{{(z - i)}^{3} {(z + i)}^{3}}$ $\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π Res (φ, i) and$ $Res (φ, i) = \lim_{z \to i} \frac{1}{(3 - 1)!} {{(z - i)}^{3} φ (z)}^{(2)}$ $= \frac{1}{2} \lim_{z \to i} {\frac{{3 z}^{2} + 1}{{(z + i)}^{3}}}^{(2)} = \frac{1}{2} {l i m}_{z \to i} {\frac{6 z {(z + i)}^{3} - 3 {(z + i)}^{2} ({3 z}^{2} + 1)}{{(z + i)}^{6}}}^{(1)}$ $= \frac{1}{2} \lim_{z \to i} {\frac{6 z (z + i) - 3 ({3 z}^{2} + 1)}{{(z + i)}^{4}}}^{(1)}$ $= \frac{1}{2} \lim_{z \to i} {\frac{- {3 z}^{2} + 6 iz - 3}{{(z + i)}^{4}}}^{(1)}$ $= \frac{1}{2} \lim_{z \to i} \frac{(- 6 z + 6 i) {(z + i)}^{4} - 4 {(z + i)}^{3} (- {3 z}^{2} + 6 iz - 3)}{{(z + i)}^{8}}$ $= \frac{1}{2} \lim_{z \to i} \frac{(- 6 z + 6 i) (z + i) - 4 (- {3 z}^{2} + 6 iz - 3)}{{(z + i)}^{5}}$ $= \frac{1}{2} . \frac{- 4 (3 - 9)}{{(2 i)}^{5}} = \frac{24}{2 . 2^{5} . i} = \frac{12}{2^{5} i} = \frac{2^{2} . 3}{2^{5} i} = \frac{3}{8 i} \Rightarrow \int_{- \infty}^{+ \infty} φ (z) dz$ $= 2 i π . \frac{3}{8 i} = \frac{3 π}{4} \Rightarrow A = \frac{8 \sqrt{3}}{27} . \frac{3 π}{4} = \frac{2 π \sqrt{3}}{9}$
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