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Question Number 125194 by bemath last updated on 09/Dec/20
∫ (((1−(√(x^2 +x+1)))^2 )/(x^2 (√(x^2 +x+1)))) dx ?
$$\:\int\:\frac{\left(\mathrm{1}−\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} \:\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}\:{dx}\:? \\ $$
Answered by liberty last updated on 09/Dec/20
by second Euler substitution we set (√(x^2 +x+1)) = xt + 1 ; then x^2 +x+1 = x^2 t^2 + 2xt + 1 ; x=((2t−1)/(1−t^2 )) dx = ((2t^2 −2t+2)/((1−t^2 )^2 )) dt ; I= ∫(((1−(√(x^2 +x+1)) )^2 )/(x^2 (√(x^2 +x+1)))) dx = 2∫ (t^2 /(1−t^2 )) dt I = −2t + ln ∣ ((1+t)/(1−t)) ∣ + c I = −((2((√(x^2 +x+1)) −1))/x) + ln ∣ ((x+(√(x^2 +x+1))−1)/(x−(√(x^2 +x+1)) +1)) ∣ + c = −((2((√(x^2 +x+1)) −1))/x) + ln ∣2x+2(√(x^2 +x+1)) +1 ∣ + c
$$\:{by}\:{second}\:{Euler}\:{substitution}\: \\ $$$$\:{we}\:{set}\:\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\:=\:{xt}\:+\:\mathrm{1}\:;\:{then}\: \\ $$$$\:{x}^{\mathrm{2}} +{x}+\mathrm{1}\:=\:{x}^{\mathrm{2}} {t}^{\mathrm{2}} \:+\:\mathrm{2}{xt}\:+\:\mathrm{1}\:;\:{x}=\frac{\mathrm{2}{t}−\mathrm{1}}{\mathrm{1}−{t}^{\mathrm{2}} } \\ $$$$\:{dx}\:=\:\frac{\mathrm{2}{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{2}}{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dt}\:;\: \\ $$$${I}=\:\int\frac{\left(\mathrm{1}−\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\:\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} \:\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}\:{dx}\:=\:\mathrm{2}\int\:\frac{{t}^{\mathrm{2}} }{\mathrm{1}−{t}^{\mathrm{2}} }\:{dt}\: \\ $$$${I}\:=\:−\mathrm{2}{t}\:+\:\mathrm{ln}\:\mid\:\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\:\mid\:+\:{c}\: \\ $$$${I}\:=\:−\frac{\mathrm{2}\left(\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\:−\mathrm{1}\right)}{{x}}\:+\:\mathrm{ln}\:\mid\:\frac{{x}+\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}−\mathrm{1}}{{x}−\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\:+\mathrm{1}}\:\mid\:+\:{c} \\ $$$$=\:−\frac{\mathrm{2}\left(\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\:−\mathrm{1}\right)}{{x}}\:+\:\mathrm{ln}\:\mid\mathrm{2}{x}+\mathrm{2}\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\:+\mathrm{1}\:\mid\:+\:{c}\: \\ $$

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