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Question Number 35428 by abdo.msup.com last updated on 18/May/18

find ∫ ((x+3)/(√(x^2 +x −1)))dx

$${find}\:\:\:\:\:\int\:\:\:\:\frac{{x}+\mathrm{3}}{\sqrt{{x}^{\mathrm{2}} \:+{x}\:−\mathrm{1}}}{dx} \\ $$

Commented by prof Abdo imad last updated on 19/May/18

let put I = ∫ ((x+3)/(√(x^2 +x−1)))dx I = (1/2) ∫ ((2x+6)/(√(x^2 +x−1)))dx =(1/2)∫ ((2x+1)/(√(x^2 +x−1)))dx +(5/2) ∫ (dx/(√(x^2 +x−1))) but ∫ ((2x+1)/(2(√(x^2 +x−1))))dx =(√(x^2 +x−1)) +λ x^2 +x−1 = x^2 +2x(1/2) +(1/4) −1−(1/4) =(x+(1/2))^2 −(5/4) and cjsngement x+(1/2) =((√5)/2)cht give ∫ (dx/(√(x^2 +x−1))) = ((√5)/2)∫ ((sht dt)/(√((5/4)(ch^2 t−1)))) = ((√5)/2) .(2/(√5)) ∫ ((sht)/(sht))dt = t +c but 2x+1=(√5) cht⇒ cht =((2x+1)/(√5)) ⇒ t =argch(((2x+1)/(√5))) =ln( ((2x+1)/(√5)) +(√({((2x+1)/(√5))}^2 −1)) ) ⇒ I = (√(x^2 +x−1)) +(5/2)ln{ ((2x+1)/(√5)) +(√((((2x+1)/(√5)))^2 −1))) +λ

$${let}\:{put}\:\:{I}\:\:=\:\int\:\:\:\:\:\:\frac{{x}+\mathrm{3}}{\sqrt{{x}^{\mathrm{2}} \:+{x}−\mathrm{1}}}{dx} \\ $$$${I}\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\:\frac{\mathrm{2}{x}+\mathrm{6}}{\sqrt{{x}^{\mathrm{2}} \:+{x}−\mathrm{1}}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\:\:\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{{x}^{\mathrm{2}} +{x}−\mathrm{1}}}{dx}\:\:\:+\frac{\mathrm{5}}{\mathrm{2}}\:\int\:\:\:\:\:\frac{{dx}}{\sqrt{{x}^{\mathrm{2}} +{x}−\mathrm{1}}}\:{but} \\ $$$$\int\:\:\:\:\:\:\:\frac{\mathrm{2}{x}+\mathrm{1}}{\mathrm{2}\sqrt{{x}^{\mathrm{2}} \:+{x}−\mathrm{1}}}{dx}\:=\sqrt{{x}^{\mathrm{2}} +{x}−\mathrm{1}}\:\:\:+\lambda \\ $$$${x}^{\mathrm{2}} \:+{x}−\mathrm{1}\:=\:{x}^{\mathrm{2}} \:+\mathrm{2}{x}\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}}\:−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$=\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:−\frac{\mathrm{5}}{\mathrm{4}}\:\:{and}\:{cjsngement}\:{x}+\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}{cht} \\ $$$${give}\:\:\int\:\:\:\:\:\frac{{dx}}{\sqrt{{x}^{\mathrm{2}} \:+{x}−\mathrm{1}}}\:=\:\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\int\:\:\:\:\:\:\frac{{sht}\:{dt}}{\sqrt{\frac{\mathrm{5}}{\mathrm{4}}\left({ch}^{\mathrm{2}} {t}−\mathrm{1}\right)}} \\ $$$$=\:\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\:.\frac{\mathrm{2}}{\sqrt{\mathrm{5}}}\:\:\int\:\:\:\:\frac{{sht}}{{sht}}{dt}\:=\:{t}\:\:\:\:+{c}\:\:{but}\:\mathrm{2}{x}+\mathrm{1}=\sqrt{\mathrm{5}}\:{cht}\Rightarrow \\ $$$${cht}\:\:=\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{\mathrm{5}}}\:\Rightarrow\:{t}\:={argch}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{\mathrm{5}}}\right) \\ $$$$={ln}\left(\:\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{\mathrm{5}}}\:+\sqrt{\left\{\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{\mathrm{5}}}\right\}^{\mathrm{2}} −\mathrm{1}}\:\right)\:\Rightarrow \\ $$$${I}\:\:=\:\sqrt{{x}^{\mathrm{2}} \:+{x}−\mathrm{1}}\:\:\:+\frac{\mathrm{5}}{\mathrm{2}}{ln}\left\{\:\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{\mathrm{5}}}\:+\sqrt{\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{\mathrm{5}}}\right)^{\mathrm{2}} \:−\mathrm{1}}\right)\:+\lambda \\ $$

Answered by ajfour last updated on 19/May/18

I=(1/2)∫((2x+1)/(√(x^2 +x−1)))dx+(5/2)∫(dx/(√((x+(1/2))^2 −(((√5)/2))^2 ))) =(√(x^2 +x−1))+(5/2)ln ∣x+(1/2)+(√(x^2 +x−1))∣+c .

$${I}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{{x}^{\mathrm{2}} +{x}−\mathrm{1}}}{dx}+\frac{\mathrm{5}}{\mathrm{2}}\int\frac{{dx}}{\sqrt{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} }} \\ $$$$=\sqrt{{x}^{\mathrm{2}} +{x}−\mathrm{1}}+\frac{\mathrm{5}}{\mathrm{2}}\mathrm{ln}\:\mid{x}+\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{{x}^{\mathrm{2}} +{x}−\mathrm{1}}\mid+{c}\:. \\ $$

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