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Question Number 29854 by abdo imad last updated on 13/Feb/18

find ∫_(−∞) ^(+∞) (((x^2 +2)dx)/(x^4 +8x^2 −16x +20)) .

$${find}\:\int_{−\infty} ^{+\infty} \:\:\:\:\:\:\:\frac{\left({x}^{\mathrm{2}} +\mathrm{2}\right){dx}}{{x}^{\mathrm{4}} \:+\mathrm{8}{x}^{\mathrm{2}} −\mathrm{16}{x}\:+\mathrm{20}}\:. \\ $$

Commented by abdo imad last updated on 18/Feb/18

we have proved that ∫_(−∞) ^(+∞) (dx/(x^2 +2ix +2−4i))=−((2π)/3) −i(π/3) but (1/(x^2 +2ix +2−4i)) = (1/(x^2 +2 +i(2x−4))) = ((x^2 +2−i(2x−4))/((x^2 +2)^2 +(2x−4)^2 ))= ((x^2 +2)/(x^4 +4x^2 +4 +4x^2 −16x +16)) −i ((2x−4)/(x^4 +4x^2 +4 +4x^2 −16x +16)) = ((x^2 +2)/(x^4 +8x^2 −16x+20)) −i ((2x−4)/(x^4 +8x^2 −16x +20)) ⇒ ∫_(−∞) ^(+∞) ((x^2 +2)/(x^4 +8x^2 −16x +20)) = ((−2π)/3) and ∫_(−∞) ^(+∞) ((2x−4)/(x^4 +8x^2 −16x +20))= (π/3) .

$${we}\:{have}\:{proved}\:{that}\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:\:+\mathrm{2}{ix}\:+\mathrm{2}−\mathrm{4}{i}}=−\frac{\mathrm{2}\pi}{\mathrm{3}}\:−{i}\frac{\pi}{\mathrm{3}}\:{but} \\ $$$$\:\:\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{2}{ix}\:+\mathrm{2}−\mathrm{4}{i}}\:=\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{2}\:+{i}\left(\mathrm{2}{x}−\mathrm{4}\right)} \\ $$$$=\:\:\:\:\frac{{x}^{\mathrm{2}} \:+\mathrm{2}−{i}\left(\mathrm{2}{x}−\mathrm{4}\right)}{\left({x}^{\mathrm{2}} \:+\mathrm{2}\right)^{\mathrm{2}} \:+\left(\mathrm{2}{x}−\mathrm{4}\right)^{\mathrm{2}} }=\:\frac{{x}^{\mathrm{2}} \:+\mathrm{2}}{{x}^{\mathrm{4}} \:+\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{4}\:+\mathrm{4}{x}^{\mathrm{2}} \:−\mathrm{16}{x}\:+\mathrm{16}} \\ $$$$\:−{i}\:\frac{\mathrm{2}{x}−\mathrm{4}}{{x}^{\mathrm{4}} \:+\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{4}\:+\mathrm{4}{x}^{\mathrm{2}} \:−\mathrm{16}{x}\:+\mathrm{16}} \\ $$$$=\:\frac{{x}^{\mathrm{2}} \:+\mathrm{2}}{{x}^{\mathrm{4}} \:+\mathrm{8}{x}^{\mathrm{2}} \:−\mathrm{16}{x}+\mathrm{20}}\:−{i}\:\frac{\mathrm{2}{x}−\mathrm{4}}{{x}^{\mathrm{4}} \:+\mathrm{8}{x}^{\mathrm{2}} \:−\mathrm{16}{x}\:+\mathrm{20}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\:\frac{{x}^{\mathrm{2}} +\mathrm{2}}{{x}^{\mathrm{4}} \:+\mathrm{8}{x}^{\mathrm{2}} \:−\mathrm{16}{x}\:+\mathrm{20}}\:=\:\frac{−\mathrm{2}\pi}{\mathrm{3}}\:{and} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\:\:\frac{\mathrm{2}{x}−\mathrm{4}}{{x}^{\mathrm{4}} \:+\mathrm{8}{x}^{\mathrm{2}} \:−\mathrm{16}{x}\:+\mathrm{20}}=\:\frac{\pi}{\mathrm{3}}\:. \\ $$

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