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Question Number 39833 by math khazana by abdo last updated on 12/Jul/18

find ∫ ((ln(x+(√(x^2 −1))))/(√(x^2 −1))) dx 2) calculate ∫_2 ^5 ((ln(x+(√(x^2 −1)))/(√(x^2 −1)))dx

$${find}\:\:\int\:\:\:\:\frac{{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} \:−\mathrm{1}}\right)}{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:{dx} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{2}} ^{\mathrm{5}} \:\:\:\frac{{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} \:−\mathrm{1}}\right.}{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{dx} \\ $$

Commented by abdo mathsup 649 cc last updated on 12/Jul/18

let I = ∫ ((ln(x+(√(x^2 −1))))/(√(x^2 −1)))dx let integrate by parts u^′ =(1/(√(x^2 −1))) and v(x)= ln(x+(√(x^2 −1))) I =ln^2 (x+(√(x^2 −1))) −∫ ((ln(x+(√(x^2 −1))))/(√(x^2 −1)))dx =ln^2 (x+(√(x^2 −1)))−I ⇒ 2I =ln^2 (x+(√(x^2 −1))) ⇒ I =(1/2)ln^2 (x+(√(x^2 −1))) 2) ∫_2 ^5 ((ln(x+(√(x^2 −1))))/(√(x^2 −1)))dx=[(1/2)ln^2 (x+(√(x^2 −1)))]_2 ^5 =(1/2){ln^2 (5 +(√(24)))−ln^2 (2+(√3))}

$${let}\:{I}\:=\:\int\:\:\frac{{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} \:−\mathrm{1}}\right)}{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{dx}\:\:{let}\:{integrate}\:{by}\:{parts} \\ $$$${u}^{'} \:=\frac{\mathrm{1}}{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:\:{and}\:\:{v}\left({x}\right)=\:{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} \:−\mathrm{1}}\right) \\ $$$${I}\:={ln}^{\mathrm{2}} \left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)\:−\int\:\:\frac{{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} \:−\mathrm{1}}\right)}{\sqrt{{x}^{\mathrm{2}} \:−\mathrm{1}}}{dx} \\ $$$$={ln}^{\mathrm{2}} \left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)−{I}\:\Rightarrow \\ $$$$\mathrm{2}{I}\:={ln}^{\mathrm{2}} \left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)\:\Rightarrow\:{I}\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left({x}+\sqrt{{x}^{\mathrm{2}} \:−\mathrm{1}}\right) \\ $$$$\left.\mathrm{2}\right)\:\int_{\mathrm{2}} ^{\mathrm{5}} \:\:\:\frac{{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)}{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{dx}=\left[\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)\right]_{\mathrm{2}} ^{\mathrm{5}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}^{\mathrm{2}} \left(\mathrm{5}\:+\sqrt{\mathrm{24}}\right)−{ln}^{\mathrm{2}} \left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\right\} \\ $$$$ \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 12/Jul/18

t=x+(√(x^2 −1)) (dt/dx)=1+((2x)/(2(√(x^2 −1))))=((x+(√(x^2 −1)))/(√(x^2 −1))) ∫lnt dt tlnt−t+c (x+(√(x^2 −1))) ln(x+(√(x^2 −1)) )−(x+(√(x^2 −1)) )+c

$${t}={x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\: \\ $$$$\frac{{dt}}{{dx}}=\mathrm{1}+\frac{\mathrm{2}{x}}{\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}=\frac{{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$\int{lnt}\:{dt} \\ $$$${tlnt}−{t}+{c} \\ $$$$\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)\:{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:\right)−\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:\right)+{c} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 12/Jul/18

t=x+(√(x^2 −1)) (dt/dx)=1+((2x)/(2(√(x^2 −1))))=((x+(√(x^2 −1)))/(√(x^2 −1))) ∫lnt dt tlnt−t+c (x+(√(x^2 −1))) ln(x+(√(x^2 −1)) )−(x+(√(x^2 −1)) )+c

$${t}={x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\: \\ $$$$\frac{{dt}}{{dx}}=\mathrm{1}+\frac{\mathrm{2}{x}}{\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}=\frac{{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$\int{lnt}\:{dt} \\ $$$${tlnt}−{t}+{c} \\ $$$$\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)\:{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:\right)−\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:\right)+{c} \\ $$

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