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Question Number 146678 by mathdanisur last updated on 14/Jul/21

∫ (√(2 + x^2 )) dx = ?

$$\int\:\sqrt{\mathrm{2}\:+\:{x}^{\mathrm{2}} }\:{dx}\:=\:? \\ $$

Answered by Ar Brandon last updated on 14/Jul/21

I=∫(√(2+x^2 ))dx =2∫(√(1+sinh^2 θ))∙coshθdθ =2∫cosh^2 θ=∫(cosh2θ+1)dθ =((sinh2θ)/2)+θ=sinhθcoshθ+θ+C =(x/( 2))(√(2+x^2 ))+argsinh((x/2))+C

$$\mathrm{I}=\int\sqrt{\mathrm{2}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$\:\:=\mathrm{2}\int\sqrt{\mathrm{1}+\mathrm{sinh}^{\mathrm{2}} \theta}\centerdot\mathrm{cosh}\theta\mathrm{d}\theta \\ $$$$\:\:=\mathrm{2}\int\mathrm{cosh}^{\mathrm{2}} \theta=\int\left(\mathrm{cosh2}\theta+\mathrm{1}\right)\mathrm{d}\theta \\ $$$$\:\:=\frac{\mathrm{sinh2}\theta}{\mathrm{2}}+\theta=\mathrm{sinh}\theta\mathrm{cosh}\theta+\theta+\mathrm{C} \\ $$$$\:\:=\frac{\mathrm{x}}{\:\mathrm{2}}\sqrt{\mathrm{2}+\mathrm{x}^{\mathrm{2}} }+\mathrm{argsinh}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)+\mathrm{C} \\ $$

Commented by mathdanisur last updated on 15/Jul/21

thankyou Ser cool

$${thankyou}\:{Ser}\:{cool} \\ $$

Answered by mathmax by abdo last updated on 14/Jul/21

I=∫(√(2+x^2 ))dx changement x=(√2)sht give I=∫ (√2)cht((√2))cht dt =∫ 2ch^2 t dt =∫ (1+ch(2t))dt =t+(1/2)sh(2t) =t +sht cht but t=argsh((x/( (√2))))=log((x/( (√2)))+(√(1+(x^2 /2)))) ⇒I=log((x/( (√2)))+(√(1+(x^2 /2)))) +(x/( (√2)))(√(1+(x^2 /2))) +c =log(x+(√(2+x^2 ))) +(x/2)(√(2+x^2 )) +C

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

Commented by mathdanisur last updated on 15/Jul/21

thankyou Ser cool

$${thankyou}\:{Ser}\:{cool} \\ $$

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