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Question Number 87686 by M±th+et£s last updated on 05/Apr/20

∫(√((ln(x+(√(1+x^2 ))))/(1+x^2 ))) dx

$$\int\sqrt{\frac{{ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)}{\mathrm{1}+{x}^{\mathrm{2}} }}\:{dx} \\ $$

Answered by TANMAY PANACEA. last updated on 05/Apr/20

t^2 =ln(x+(√(1+x^2 )) ) 2t(dt/dx)=(1/(x+(√(1+x^2 ))))×(1+((2x)/(2(√(1+x^2 )))))=(1/(√(1+x^2 ))) 2tdt=(dx/(√(1+x^2 ))) ∫(√(ln(x+(√(1+x^2 )) )) ×(dx/(√(1+x^2 ))) ∫t×2tdt 2×(t^3 /3)+c (2/3){ln(x+(√(1+x^2 )) )}^(3/2) +c

$${t}^{\mathrm{2}} ={ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\right) \\ $$$$\mathrm{2}{t}\frac{{dt}}{{dx}}=\frac{\mathrm{1}}{{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}×\left(\mathrm{1}+\frac{\mathrm{2}{x}}{\mathrm{2}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right)=\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$$\mathrm{2}{tdt}=\frac{{dx}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$$\int\sqrt{{ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\right.}\:×\frac{{dx}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$$\int{t}×\mathrm{2}{tdt} \\ $$$$\mathrm{2}×\frac{{t}^{\mathrm{3}} }{\mathrm{3}}+{c} \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}\left\{{ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\right)\right\}^{\frac{\mathrm{3}}{\mathrm{2}}} +{c} \\ $$

Answered by Ar Brandon last updated on 05/Apr/20

I=∫(√((ln(x+(√(1+x^2 ))))/(1+x^2 ))) dx Let x=tanθ⇒dx=sec^2 θdθ ⇒I=∫(√(ln(sec θ+tan θ))) sec θdθ ⇒I=(√(ln(sec θ+tan θ)))∫sec θ−∫{(d/dθ)(√(ln(sec θ+tan θ)))∫sec θ}dθ ⇒I=(ln(sec θ+tan θ))^(3/2) −∫{((sec θ)/(2(√(ln(sec θ+tan θ)))))∙ln(sec θ+tan θ)}dθ ⇒I=(ln(sec θ+tan θ))^(3/2) −(1/2)∫(√(ln(sec θ+tan θ))) secdθ ⇒(3/2)I=(ln(sec θ+tan θ))^(3/2) ⇒I=(2/3)(ln(sec θ+tan θ))^(3/2) ⇒∫(√((ln(x+(√(1+x^2 ))))/(1+x^2 ))) dx=(2/3)(ln(sec(tan^(−1) x)+x))^(3/2) +C C∈R

$$\boldsymbol{{I}}=\int\sqrt{\frac{{ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)}{\mathrm{1}+{x}^{\mathrm{2}} }}\:{dx} \\ $$$${Let}\:{x}={tan}\theta\Rightarrow{dx}={sec}^{\mathrm{2}} \theta{d}\theta \\ $$$$\Rightarrow\boldsymbol{{I}}=\int\sqrt{{ln}\left({sec}\:\theta+{tan}\:\theta\right)}\:{sec}\:\theta{d}\theta \\ $$$$\Rightarrow\boldsymbol{{I}}=\sqrt{{ln}\left({sec}\:\theta+{tan}\:\theta\right)}\int{sec}\:\theta−\int\left\{\frac{{d}}{{d}\theta}\sqrt{{ln}\left({sec}\:\theta+{tan}\:\theta\right)}\int{sec}\:\theta\right\}{d}\theta \\ $$$$\Rightarrow\boldsymbol{{I}}=\left({ln}\left({sec}\:\theta+{tan}\:\theta\right)\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\int\left\{\frac{{sec}\:\theta}{\mathrm{2}\sqrt{{ln}\left({sec}\:\theta+{tan}\:\theta\right)}}\centerdot{ln}\left({sec}\:\theta+{tan}\:\theta\right)\right\}{d}\theta \\ $$$$\Rightarrow\boldsymbol{{I}}=\left({ln}\left({sec}\:\theta+{tan}\:\theta\right)\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\frac{\mathrm{1}}{\mathrm{2}}\int\sqrt{{ln}\left({sec}\:\theta+{tan}\:\theta\right)}\:{secd}\theta \\ $$$$\Rightarrow\frac{\mathrm{3}}{\mathrm{2}}\boldsymbol{{I}}=\left({ln}\left({sec}\:\theta+{tan}\:\theta\right)\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$\Rightarrow\boldsymbol{{I}}=\frac{\mathrm{2}}{\mathrm{3}}\left({ln}\left({sec}\:\theta+{tan}\:\theta\right)\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$\Rightarrow\int\sqrt{\frac{{ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)}{\mathrm{1}+{x}^{\mathrm{2}} }}\:{dx}=\frac{\mathrm{2}}{\mathrm{3}}\left({ln}\left({sec}\left({tan}^{−\mathrm{1}} {x}\right)+{x}\right)\right)^{\frac{\mathrm{3}}{\mathrm{2}}} +{C} \\ $$$${C}\in\mathbb{R} \\ $$

Commented by M±th+et£s last updated on 05/Apr/20

god bless you

$${god}\:{bless}\:{you} \\ $$

Commented by Ar Brandon last updated on 05/Apr/20

*G. ��

Answered by mind is power last updated on 05/Apr/20

x=sh(t)⇒ =∫(√((ln(sh(t)+ch(t)))/(1+sh^2 (t)))).ch(t)dt =∫(√((ln(e^t ))/(ch^2 (t)))).ch(t)dt=∫(√t)dt=(3/2)t^(3/2) +c =(3/2)argsh^(3/2) (x)+c

$${x}={sh}\left({t}\right)\Rightarrow \\ $$$$=\int\sqrt{\frac{{ln}\left({sh}\left({t}\right)+{ch}\left({t}\right)\right)}{\mathrm{1}+{sh}^{\mathrm{2}} \left({t}\right)}}.{ch}\left({t}\right){dt} \\ $$$$=\int\sqrt{\frac{{ln}\left({e}^{{t}} \right)}{{ch}^{\mathrm{2}} \left({t}\right)}}.{ch}\left({t}\right){dt}=\int\sqrt{{t}}{dt}=\frac{\mathrm{3}}{\mathrm{2}}{t}^{\frac{\mathrm{3}}{\mathrm{2}}} +{c} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}{argsh}^{\frac{\mathrm{3}}{\mathrm{2}}} \left({x}\right)+{c} \\ $$$$ \\ $$

Commented by M±th+et£s last updated on 05/Apr/20

god bless you sir

$${god}\:{bless}\:{you}\:{sir} \\ $$

Commented by Ar Brandon last updated on 05/Apr/20

I think it should be (2/3)

$${I}\:{think}\:{it}\:{should}\:{be}\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$

Commented by M±th+et£s last updated on 05/Apr/20

yes i think its typo

$${yes}\:{i}\:{think}\:{its}\:{typo}\: \\ $$

Commented by Ar Brandon last updated on 05/Apr/20

Nice initiative though. Bravo!

$${Nice}\:{initiative}\:{though}.\:{Bravo}! \\ $$

Commented by mind is power last updated on 06/Apr/20

yeah sorrh (2/3)

$${yeah}\:{sorrh}\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$

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