Question-92723

Question Number 92723 by Power last updated on 08/May/20

Answered by Rio Michael last updated on 08/May/20

∫(dx/( (√(x^2 −4)))) let x = 2 sec u ⇒ (dx/du) = 2 sec u tan u ⇒ dx = 2 sec u tan u du ∫(dx/( (√(x^2 −4)))) = ∫((2 sec u tan u du)/( (√((2 sec u)^2 − 4)))) = ∫((2 sec u tan u du)/( (√(4 sec^2 u −4)))) = ∫((sec u tan u du)/(tan u)) = ∫ sec u du = ln∣ sec u + tan u∣ + C do all replacements and see that ∫ (dx/( (√(x^2 −4)))) = ln∣ x + (√(x^2 −4)) ∣ + C

$$\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}}\:\mathrm{let}\:{x}\:=\:\mathrm{2}\:\mathrm{sec}\:{u} \\ $$$$\Rightarrow\:\frac{{dx}}{{du}}\:=\:\mathrm{2}\:\mathrm{sec}\:{u}\:\mathrm{tan}\:{u} \\ $$$$\Rightarrow\:{dx}\:=\:\mathrm{2}\:\mathrm{sec}\:{u}\:\mathrm{tan}\:{u}\:{du} \\ $$$$\:\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}}\:=\:\int\frac{\mathrm{2}\:\mathrm{sec}\:{u}\:\mathrm{tan}\:{u}\:{du}}{\:\sqrt{\left(\mathrm{2}\:\mathrm{sec}\:{u}\right)^{\mathrm{2}} −\:\mathrm{4}}}\:=\:\int\frac{\mathrm{2}\:\mathrm{sec}\:{u}\:\mathrm{tan}\:{u}\:{du}}{\:\sqrt{\mathrm{4}\:\mathrm{sec}^{\mathrm{2}} {u}\:−\mathrm{4}}} \\ $$$$\:=\:\int\frac{\mathrm{sec}\:{u}\:\mathrm{tan}\:{u}\:{du}}{\mathrm{tan}\:{u}}\:\:=\:\int\:\mathrm{sec}\:{u}\:{du}\:=\:\mathrm{ln}\mid\:\mathrm{sec}\:{u}\:+\:\mathrm{tan}\:{u}\mid\:+\:{C} \\ $$$$\mathrm{do}\:\mathrm{all}\:\mathrm{replacements}\:\mathrm{and}\:\mathrm{see}\:\mathrm{that} \\ $$$$\int\:\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}}\:=\:\mathrm{ln}\mid\:{x}\:+\:\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}\:\mid\:+\:{C} \\ $$

Commented by mathmax by abdo last updated on 08/May/20

I=∫ (dx/( (√(x^2 −4)))) changement x =2ch(t) give I =∫ ((2sh(t))/(2sht))dt =2t +c but t =argch((x/2))=ln((x/2)+(√((x^2 /4)−1))) =ln(x+(√(x^2 −4)))−ln(2) ⇒ I =2ln(x+(√(x^2 −4))) +C

$${I}=\int\:\:\:\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}}\:\:{changement}\:{x}\:=\mathrm{2}{ch}\left({t}\right)\:{give} \\ $$$${I}\:=\int\:\frac{\mathrm{2}{sh}\left({t}\right)}{\mathrm{2}{sht}}{dt}\:=\mathrm{2}{t}\:+{c}\:\:{but}\:{t}\:={argch}\left(\frac{{x}}{\mathrm{2}}\right)={ln}\left(\frac{{x}}{\mathrm{2}}+\sqrt{\frac{{x}^{\mathrm{2}} }{\mathrm{4}}−\mathrm{1}}\right) \\ $$$$={ln}\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}\right)−{ln}\left(\mathrm{2}\right)\:\:\Rightarrow \\ $$$${I}\:=\mathrm{2}{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}\right)\:+{C} \\ $$

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