find ∫_(−∞) ^(+∞) ((x^2 −x)/(x^4 −x^2 +3))dx


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Question Number 78267 by msup trace by abdo last updated on 15/Jan/20

$f i n d \int_{- \infty}^{+ \infty} \frac{x^{2} - x}{x^{4} - x^{2} + 3} d x$
Commented by mathmax by abdo last updated on 17/Jan/20
$let A =∫_(−∞) ^(+∞) ((x^2 −x)/(x^4 −x^2 +3))dx ⇒ A =∫_(−∞) ^(+∞) (x^2 /(x^4 −x^2 +3))dx−∫_(−∞) ^(+∞) ((xdx)/(x^4 −x^2 +3))(→odd) =∫_(−∞) ^(+∞) (x^2 /(x^4 −x^2 +3))dx let W(z)=(z^2 /(z^4 −z^2 +3)) poles of W? z^4 −z^2 +3 =0 ⇒t^2 −t+3 =0 (t=z^2 ) Δ=1−12 =−11 ⇒t_1 =((1+i(√(11)))/2) and t_2 =((1−i(√(11)))/2) ∣t_1 ∣ =(1/2)(√(1+11))=(1/2)(2(√3)) =(√3) ⇒t_1 =(√3)e^(iarctan((√(11)))) t_2 =conj(t_1 )=(√3)e^(−iarctan((√(11)))) ⇒W(z)=(z^2 /((z^2 −(√3)e^(iarctan((√(11)))) (z^2 −(√3)e^(−iarctan((√(11)))) ))) =(z^2 /((z−α e^((i/2)arctan((√(11)))) )(z+αe^((i/2)arctan((√(11)))) )(z−αe^(−(i/2)arctan((√(11)))) )(z+αe^(−(i/2)arctan((√(11)))) ))) (with α =(√(√3))) ∫_(−∞) ^(+∞) W(z)dz =2iπ{ Res(W,αe^((i/2)arctan((√(11)))) )+Res(W,−αe^(−(i/2)arctan((√(11)))) )} Res(W,αe^((i/2)arctan((√(11)))) ) =((α^2 e^(iarctan((√(11)))) )/(2αe^((i/2)arctan((√(11)))) (α^2 e^(iarctan((√(11)))) −α^2 e^(−iarctan((√(11)))) ))) =(1/(2α))×(e^((i/2)arctan((√(11)))) /(2i sin(arctan((√(11)))))) =(1/(4iαsin(arctan((√(11)))))×e^((i/2)arctan((√(11)))) ....be continued...$
$l e t A = \int_{- \infty}^{+ \infty} \frac{x^{2} - x}{x^{4} - x^{2} + 3} d x \Rightarrow A = \int_{- \infty}^{+ \infty} \frac{x^{2}}{x^{4} - x^{2} + 3} d x - \int_{- \infty}^{+ \infty} \frac{x d x}{x^{4} - x^{2} + 3} (\to o d d)$ $= \int_{- \infty}^{+ \infty} \frac{x^{2}}{x^{4} - x^{2} + 3} d x l e t W (z) = \frac{z^{2}}{z^{4} - z^{2} + 3} p o l e s o f W ?$ $z^{4} - z^{2} + 3 = 0 \Rightarrow t^{2} - t + 3 = 0 (t = z^{2})$ $Δ = 1 - 12 = - 11 \Rightarrow t_{1} = \frac{1 + i \sqrt{11}}{2} a n d t_{2} = \frac{1 - i \sqrt{11}}{2}$ $∣ t_{1} ∣ = \frac{1}{2} \sqrt{1 + 11} = \frac{1}{2} (2 \sqrt{3}) = \sqrt{3} \Rightarrow t_{1} = \sqrt{3} e^{i a r c t a n (\sqrt{11})}$ $t_{2} = c o n j (t_{1}) = \sqrt{3} e^{- i a r c t a n (\sqrt{11})} \Rightarrow W (z) = \frac{z^{2}}{(z^{2} - \sqrt{3} e^{i a r c t a n (\sqrt{11})} (z^{2} - \sqrt{3} e^{- i a r c t a n (\sqrt{11})})}$ $= \frac{z^{2}}{(z - α e^{\frac{i}{2} a r c t a n (\sqrt{11})}) (z + α e^{\frac{i}{2} a r c t a n (\sqrt{11})}) (z - α e^{- \frac{i}{2} a r c t a n (\sqrt{11})}) (z + α e^{- \frac{i}{2} a r c t a n (\sqrt{11})})}$ $(w i t h α = \sqrt{\sqrt{3}})$ $\int_{- \infty}^{+ \infty} W (z) d z = 2 i π {R e s (W, α e^{\frac{i}{2} a r c t a n (\sqrt{11})}) + R e s (W, - α e^{- \frac{i}{2} a r c t a n (\sqrt{11})})}$ $R e s (W, α e^{\frac{i}{2} a r c t a n (\sqrt{11})}) = \frac{α^{2} e^{i a r c t a n (\sqrt{11})}}{2 α e^{\frac{i}{2} a r c t a n (\sqrt{11})} (α^{2} e^{i a r c t a n (\sqrt{11})} - α^{2} e^{- i a r c t a n (\sqrt{11})})}$ $= \frac{1}{2 α} \times \frac{e^{\frac{i}{2} a r c t a n (\sqrt{11})}}{2 i s i n (a r c t a n (\sqrt{11}))} = \frac{1}{4 i α s i n (a r c t a n (\sqrt{11})} \times e^{\frac{i}{2} a r c t a n (\sqrt{11})}$ $. . . . b e c o n t i n u e d . . .$
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