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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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