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


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Question Number 56931 by turbo msup by abdo last updated on 26/Mar/19

$c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{x^{2} - 1}{{(x^{2} - x + 3)}^{2}} d x$
Commented by maxmathsup by imad last updated on 27/Mar/19
$residus method let ϕ(z) =((z^2 −1)/((z^2 −z +3)^2 )) let determine the poles of ϕ? z^2 −z+3=0 →Δ =1−12 =−11=(i(√(11)))^2 ⇒z_1 =((1+i(√(11)))/2) z_2 =((1−i(√(11)))/2) the poles of ϕ are z_1 and z_2 (doubles) and ϕ(z)=((z^2 −1)/((z−z_1 )^2 (z−z_2 )^2 )) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπRes(ϕ,z_1 ) Res(ϕ,z_1 ) = lim_(z→z_1 ) (1/((2−1)!)) {(z−z_1 )^2 ϕ(z)}^((1)) =lim_(z→z_1 ) {((z^2 −1)/((z−z_2 )^2 ))}^((1)) =lim_(z→z_1 ) {((2z(z−z_2 )^2 −2(z−z_2 )(z^2 −1))/((z−z_2 )^4 ))} =lim_(z→z_1 ) ((2z(z−z_2 )−2(z^2 −1))/((z−z_2 )^3 )) =((2z_1 (z_1 −z_2 )−2(z_1 ^2 −1))/((z_1 −z_2 )^3 )) =((2i(√(11))(((1+i(√(11)))/2))−2( (1/4)(1+2i(√(11))−11)−1))/((i(√(11)))^3 )) =((i(√(11))(1+i(√(11)))−2{((i(√(11)))/2)−(5/2)−1})/(−11i(√(11)))) =((i(√(11))−11−i(√(11))+5 +2)/(−11i(√(11)))) =((−4)/(−11i(√(11)))) =(4/(11i(√(11)))) ⇒∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (4/(11i(√(11)))) =((8π)/(11(√(11)))) =((8π(√(11)))/(121)) ★∫_(−∞) ^(+∞) ((x^2 −1)/((x^2 −x+3)^2 )) dx =((8π(√(11)))/(121)) ★$
$r e s i d u s m e t h o d l e t φ (z) = \frac{z^{2} - 1}{{(z^{2} - z + 3)}^{2}} l e t d e t e r m i n e t h e p o l e s o f φ ?$ $z^{2} - z + 3 = 0 \to Δ = 1 - 12 = - 11 = {(i \sqrt{11})}^{2} \Rightarrow z_{1} = \frac{1 + i \sqrt{11}}{2}$ $z_{2} = \frac{1 - i \sqrt{11}}{2} t h e p o l e s o f φ a r e z_{1} a n d z_{2} (d o u b l e s) a n d φ (z) = \frac{z^{2} - 1}{{(z - z_{1})}^{2} {(z - z_{2})}^{2}}$ $r e s i d u s t h e o r e m g i v e \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, z_{1})$ $R e s (φ, z_{1}) = {l i m}_{z \to z_{1}} \frac{1}{(2 - 1)!} {{(z - z_{1})}^{2} φ (z)}^{(1)}$ $= {l i m}_{z \to z_{1}} {\frac{z^{2} - 1}{{(z - z_{2})}^{2}}}^{(1)} = {l i m}_{z \to z_{1}} {\frac{2 z {(z - z_{2})}^{2} - 2 (z - z_{2}) (z^{2} - 1)}{{(z - z_{2})}^{4}}}$ $= {l i m}_{z \to z_{1}} \frac{2 z (z - z_{2}) - 2 (z^{2} - 1)}{{(z - z_{2})}^{3}} = \frac{2 z_{1} (z_{1} - z_{2}) - 2 (z_{1}^{2} - 1)}{{(z_{1} - z_{2})}^{3}}$ $= \frac{2 i \sqrt{11} (\frac{1 + i \sqrt{11}}{2}) - 2 (\frac{1}{4} (1 + 2 i \sqrt{11} - 11) - 1)}{{(i \sqrt{11})}^{3}}$ $= \frac{i \sqrt{11} (1 + i \sqrt{11}) - 2 {\frac{i \sqrt{11}}{2} - \frac{5}{2} - 1}}{- 11 i \sqrt{11}} = \frac{i \sqrt{11} - 11 - i \sqrt{11} + 5 + 2}{- 11 i \sqrt{11}}$ $= \frac{- 4}{- 11 i \sqrt{11}} = \frac{4}{11 i \sqrt{11}} \Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{4}{11 i \sqrt{11}} = \frac{8 π}{11 \sqrt{11}} = \frac{8 π \sqrt{11}}{121}$ $★ \int_{- \infty}^{+ \infty} \frac{x^{2} - 1}{{(x^{2} - x + 3)}^{2}} d x = \frac{8 π \sqrt{11}}{121} ★$
	Answered by MJS last updated on 27/Mar/19

	${Ostrogradski}^{'} s Method once again$ $\int \frac{p (x)}{q (x)} d x = \frac{p_{1} (x)}{q_{1} (x)} + \int \frac{p_{2} (x)}{q_{2} (x)} d x$ $q_{1} (x) = \gcd (q (x), q^{'} (x))$ $q_{2} (x) = \frac{q (x)}{q_{1} (x)}$ $p_{1} (x), p_{2} (x) can be found like this :$ $\frac{p (x)}{q (x)} = \frac{d}{d x} [\frac{p_{1} (x)}{q_{1} (x)}] + \frac{p_{2} (x)}{q_{2} (x)}$ $in our case :$ $\int \frac{x^{2} - 1}{{(x^{2} - x + 3)}^{2}} d x = - \frac{7 x + 2}{11 (x^{2} - x + 3)} + \frac{4}{11} \int \frac{d x}{x^{2} - x + 3} =$ $= - \frac{7 x + 2}{11 (x^{2} - x + 3)} + \frac{8 \sqrt{11}}{121} \arctan (\frac{\sqrt{11}}{11} (2 x - 1)) + C$ $\underset{- \infty}{\int^{+ \infty}} \frac{x^{2} - 1}{{(x^{2} - x + 3)}^{2}} d x = \frac{8 \sqrt{11}}{121} π$
	Commented by turbo msup by abdo last updated on 27/Mar/19

	$t h a n k s s i r m j s .$
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