x-2-4-dx-answer-quick-pls-

Question Number 86626 by sakeefhasan05@gmail.com last updated on 29/Mar/20

$$\int\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{4}}\:\:\mathrm{dx} \\ $$$$\mathrm{answer}\:\mathrm{quick}\:\mathrm{pls} \\ $$

Commented by sakeefhasan05@gmail.com last updated on 30/Mar/20

Commented by mathmax by abdo last updated on 30/Mar/20

let A =∫(√(x^2 +4))dx changement x =2sh(t)give A =∫ 2ch(t)2ch(t)dt =4 ∫ ch^2 t dt =2∫(1+ch(2t))dt =2t +sh(2t)+C =2t +2sh(t)ch(t) +C =2argsh((x/2))+ x(√(1+(x^2 /4))) +C =2ln((x/2)+(√(1+(x^2 /4))))+(x/2)(√(x^2 +4))+C =2ln(x+(√(x^2 +4))) +(1/2)x(√(x^2 +4)) +c

$${let}\:{A}\:=\int\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}{dx}\:{changement}\:{x}\:=\mathrm{2}{sh}\left({t}\right){give} \\ $$$${A}\:=\int\:\mathrm{2}{ch}\left({t}\right)\mathrm{2}{ch}\left({t}\right){dt}\:=\mathrm{4}\:\int\:{ch}^{\mathrm{2}} {t}\:{dt}\:=\mathrm{2}\int\left(\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)\right){dt} \\ $$$$=\mathrm{2}{t}\:+{sh}\left(\mathrm{2}{t}\right)+{C}\:=\mathrm{2}{t}\:+\mathrm{2}{sh}\left({t}\right){ch}\left({t}\right)\:+{C} \\ $$$$=\mathrm{2}{argsh}\left(\frac{{x}}{\mathrm{2}}\right)+\:{x}\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{4}}}\:\:+{C} \\ $$$$=\mathrm{2}{ln}\left(\frac{{x}}{\mathrm{2}}+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{4}}}\right)+\frac{{x}}{\mathrm{2}}\sqrt{{x}^{\mathrm{2}} \:+\mathrm{4}}+{C} \\ $$$$=\mathrm{2}{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}\:+{c} \\ $$

Answered by TANMAY PANACEA. last updated on 29/Mar/20

(√(x^2 +4)) ∫dx−∫[((d((√(x^2 +4)) ))/dx)∫dx]dx x(√(x^2 +4)) −∫((2x×x)/(2(√(x^2 +4))))dx x(√(x^2 +4)) −∫((x^2 +4−4)/( (√(x^2 +4)) ))dx x(√(x^2 +4)) −∫(√(x^2 +4)) +4∫(dx/( (√(x^2 +4)))) 2I=x(√(x^2 +4)) +4ln(x+(√(x^2 +4 ))) I=((x(√(x^2 +4)))/2)+2ln(x+(√(x^2 +4)) )

$$\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}\:\int{dx}−\int\left[\frac{{d}\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}\:\right)}{{dx}}\int{dx}\right]{dx} \\ $$$${x}\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}\:−\int\frac{\mathrm{2}{x}×{x}}{\mathrm{2}\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}}{dx} \\ $$$${x}\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}\:−\int\frac{{x}^{\mathrm{2}} +\mathrm{4}−\mathrm{4}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}\:}{dx} \\ $$$${x}\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}\:−\int\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}\:+\mathrm{4}\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}} \\ $$$$\mathrm{2}{I}={x}\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}\:+\mathrm{4}{ln}\left({x}+\sqrt{\left.{x}^{\mathrm{2}} +\mathrm{4}\:\right)}\right. \\ $$$${I}=\frac{{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}+\mathrm{2}{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}\:\right) \\ $$

Commented by sakeefhasan05@gmail.com last updated on 29/Mar/20

thank u very much can u pls do it again with another method (x=2tan θ) substition

$$\mathrm{thank}\:\mathrm{u}\:\mathrm{very}\:\mathrm{much}\: \\ $$$$\mathrm{can}\:\mathrm{u}\:\mathrm{pls}\:\mathrm{do}\:\mathrm{it}\:\mathrm{again}\:\mathrm{with}\:\mathrm{another}\: \\ $$$$\mathrm{method}\:\:\left(\mathrm{x}=\mathrm{2tan}\:\theta\right)\:\mathrm{substition} \\ $$

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