Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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

$$\:{by}\:{second}\:{Euler}\:{substitution}\: \\ $$$$\:{we}\:{set}\:\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\:=\:{xt}\:+\:\mathrm{1}\:;\:{then}\: \\ $$$$\:{x}^{\mathrm{2}} +{x}+\mathrm{1}\:=\:{x}^{\mathrm{2}} {t}^{\mathrm{2}} \:+\:\mathrm{2}{xt}\:+\:\mathrm{1}\:;\:{x}=\frac{\mathrm{2}{t}−\mathrm{1}}{\mathrm{1}−{t}^{\mathrm{2}} } \\ $$$$\:{dx}\:=\:\frac{\mathrm{2}{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{2}}{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dt}\:;\: \\ $$$${I}=\:\int\frac{\left(\mathrm{1}−\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\:\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} \:\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}\:{dx}\:=\:\mathrm{2}\int\:\frac{{t}^{\mathrm{2}} }{\mathrm{1}−{t}^{\mathrm{2}} }\:{dt}\: \\ $$$${I}\:=\:−\mathrm{2}{t}\:+\:\mathrm{ln}\:\mid\:\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\:\mid\:+\:{c}\: \\ $$$${I}\:=\:−\frac{\mathrm{2}\left(\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\:−\mathrm{1}\right)}{{x}}\:+\:\mathrm{ln}\:\mid\:\frac{{x}+\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}−\mathrm{1}}{{x}−\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\:+\mathrm{1}}\:\mid\:+\:{c} \\ $$$$=\:−\frac{\mathrm{2}\left(\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\:−\mathrm{1}\right)}{{x}}\:+\:\mathrm{ln}\:\mid\mathrm{2}{x}+\mathrm{2}\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\:+\mathrm{1}\:\mid\:+\:{c}\: \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com