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Question Number 65354 by mathmax by abdo last updated on 28/Jul/19

find ∫ (dx/(√(x^2 +x−2)))

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

Commented by mathmax by abdo last updated on 29/Jul/19

let A =∫ (dx/(√(x^2 +x−2))) we have x^2 +x−2 =x^2 −1 +x−1 =(x−1)(x+1)+x−1 =(x−1)(x+2) ⇒A =∫ (dx/((√(x−1))(√(x−2)))) changement (√(x−1))=t give x−1 =t^2 ⇒x =t^2 +1 ⇒ A =∫ ((2tdt)/(t(√(t^2 −1)))) =2 ∫ (dt/(√(t^2 −1))) =2ln(t+(√(t^2 −1))) +c =2ln((√(x−1))+(√(x−2))) +c

$${let}\:{A}\:=\int\:\:\:\:\:\frac{{dx}}{\sqrt{{x}^{\mathrm{2}} \:+{x}−\mathrm{2}}}\:\:{we}\:{have}\:\:{x}^{\mathrm{2}} +{x}−\mathrm{2}\:={x}^{\mathrm{2}} −\mathrm{1}\:+{x}−\mathrm{1} \\ $$$$=\left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)+{x}−\mathrm{1}\:=\left({x}−\mathrm{1}\right)\left({x}+\mathrm{2}\right)\:\Rightarrow{A}\:=\int\:\:\:\frac{{dx}}{\sqrt{{x}−\mathrm{1}}\sqrt{{x}−\mathrm{2}}} \\ $$$${changement}\:\sqrt{{x}−\mathrm{1}}={t}\:{give}\:{x}−\mathrm{1}\:={t}^{\mathrm{2}} \:\Rightarrow{x}\:={t}^{\mathrm{2}} \:+\mathrm{1}\:\Rightarrow \\ $$$${A}\:=\int\:\:\:\:\frac{\mathrm{2}{tdt}}{{t}\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}}\:\:=\mathrm{2}\:\int\:\frac{{dt}}{\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}}\:=\mathrm{2}{ln}\left({t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\right)\:+{c} \\ $$$$=\mathrm{2}{ln}\left(\sqrt{{x}−\mathrm{1}}+\sqrt{{x}−\mathrm{2}}\right)\:+{c}\:\: \\ $$

Commented by MJS last updated on 29/Jul/19

typo at the end 2ln ((√(x−1))+(√(x+2))) +C

$$\mathrm{typo}\:\mathrm{at}\:\mathrm{the}\:\mathrm{end} \\ $$$$\mathrm{2ln}\:\left(\sqrt{{x}−\mathrm{1}}+\sqrt{{x}+\mathrm{2}}\right)\:+{C} \\ $$

Commented by mathmax by abdo last updated on 30/Jul/19

thank you sir MJS

$${thank}\:{you}\:{sir}\:{MJS} \\ $$

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