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Question Number 93193 by john santu last updated on 11/May/20
∫ ((ln(x+(√(x^2 −1))))/( (√((x^2 −1)^3 )))) dx ?
$$\int\:\frac{\mathrm{ln}\left(\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\right)}{\:\sqrt{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}} }}\:\mathrm{dx}\:?\: \\ $$
Commented by abdomathmax last updated on 14/May/20
I =∫ ((ln(x+(√(x^2 −1))))/( (√((x^2 −1)^3 ))))dx vhangement x=cht give I =∫ ((ln(cht +sht))/( (√(sh^6 t)))) sh(t)dt =∫ (t/(sh^2 t)) dt =∫ ((2t)/(ch(2t)−1))dt =_(2t =u) ∫ (u/(ch(u)−1))(du/2) =(1/2)∫ (u/(((e^u +e^(−u) )/2)−1))du =∫ (u/(e^u +e^(−u) −2)) du =_(e^u =z) ∫ ((ln(z))/(z+z^(−1) −2))×(dz/z) =∫ ((lnz)/(z^2 −2z +1))dz =∫ ((lnz)/((z−1)^2 ))dx =−(1/(z−1))ln(z)+∫ (1/(z−1))×(dz/z) =−(1/(z−1))lnz +∫((1/(z−1))−(1/z))dz =−(1/(z−1))lnz +ln∣((z−1)/z)∣ +C =−(u/(e^u −1))+ln∣((e^u −1)/e^u )∣ +C u =(t/2) =(1/2)argch(x) =(1/2)ln(x+(√(x^2 −1))) ⇒ e^u =(√(x+(√(x^2 −1)))) ⇒ I =−((ln(x+(√(x^2 −1))))/(2((√(x+(√(x^2 −1))))))) +ln∣(((√(x+(√(x^2 −1))))−1)/( (√(x+(√(x^2 −1))))))∣ +C
$${I}\:=\int\:\:\frac{{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)}{\:\sqrt{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}} }}{dx}\:\:{vhangement}\:{x}={cht}\:{give} \\ $$$${I}\:=\int\:\:\frac{{ln}\left({cht}\:+{sht}\right)}{\:\sqrt{{sh}^{\mathrm{6}} {t}}}\:{sh}\left({t}\right){dt} \\ $$$$=\int\:\:\:\:\:\frac{{t}}{{sh}^{\mathrm{2}} {t}}\:{dt}\:\:=\int\:\:\:\frac{\mathrm{2}{t}}{{ch}\left(\mathrm{2}{t}\right)−\mathrm{1}}{dt} \\ $$$$=_{\mathrm{2}{t}\:={u}} \:\:\:\int\:\:\:\:\frac{{u}}{{ch}\left({u}\right)−\mathrm{1}}\frac{{du}}{\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\:\frac{{u}}{\frac{{e}^{{u}} +{e}^{−{u}} }{\mathrm{2}}−\mathrm{1}}{du} \\ $$$$=\int\:\:\:\:\:\frac{{u}}{{e}^{{u}} \:+{e}^{−{u}} −\mathrm{2}}\:{du}\:\:\:=_{{e}^{{u}} ={z}} \:\:\:\:\int\:\:\:\frac{{ln}\left({z}\right)}{{z}+{z}^{−\mathrm{1}} −\mathrm{2}}×\frac{{dz}}{{z}} \\ $$$$=\int\:\:\:\:\:\:\:\frac{{lnz}}{{z}^{\mathrm{2}} −\mathrm{2}{z}\:+\mathrm{1}}{dz}\:=\int\:\:\frac{{lnz}}{\left({z}−\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$=−\frac{\mathrm{1}}{{z}−\mathrm{1}}{ln}\left({z}\right)+\int\:\:\frac{\mathrm{1}}{{z}−\mathrm{1}}×\frac{{dz}}{{z}} \\ $$$$=−\frac{\mathrm{1}}{{z}−\mathrm{1}}{lnz}\:+\int\left(\frac{\mathrm{1}}{{z}−\mathrm{1}}−\frac{\mathrm{1}}{{z}}\right){dz} \\ $$$$=−\frac{\mathrm{1}}{{z}−\mathrm{1}}{lnz}\:+{ln}\mid\frac{{z}−\mathrm{1}}{{z}}\mid\:+{C} \\ $$$$=−\frac{{u}}{{e}^{{u}} −\mathrm{1}}+{ln}\mid\frac{{e}^{{u}} −\mathrm{1}}{{e}^{{u}} }\mid\:+{C} \\ $$$${u}\:=\frac{{t}}{\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{2}}{argch}\left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)\:\Rightarrow \\ $$$${e}^{{u}} \:=\sqrt{{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:\Rightarrow \\ $$$${I}\:=−\frac{{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)}{\mathrm{2}\left(\sqrt{\left.{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)}\right.}\:+{ln}\mid\frac{\sqrt{{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}−\mathrm{1}}{\:\sqrt{{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}}\mid\:+{C} \\ $$
Answered by MJS last updated on 11/May/20
∫((ln (x+(√(x^2 −1))))/((x^2 −1)^(3/2) ))dx= [t=x+(√(x^2 −1)) → dx=((√(x^2 −1))/(x+(√(x^2 −1))))dt] =4∫((t ln t)/((t^2 −1)^2 ))dt= by parts u=ln t → u′=(1/t) v′=((4t)/((t^2 −1)^2 )) → v=−(2/(t^2 −1)) =−((2ln t)/(t^2 −1))+2∫(dt/(t(t^2 −1)))= =−((2ln t)/(t^2 −1))+ln (t^2 −1) −2ln t = =−((2t^2 )/(t^2 −1))ln t +ln (t^2 −1) = =−((x+(√(x^2 −1)))/( (√(x^2 −1))))ln (x+(√(x^2 −1))) +ln (x^2 −1+x(√(x^2 −1))) +C
$$\int\frac{\mathrm{ln}\:\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} }{dx}= \\ $$$$\:\:\:\:\:\left[{t}={x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:\rightarrow\:{dx}=\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{dt}\right] \\ $$$$=\mathrm{4}\int\frac{{t}\:\mathrm{ln}\:{t}}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }{dt}= \\ $$$$\:\:\:\:\:\mathrm{by}\:\mathrm{parts} \\ $$$$\:\:\:\:\:{u}=\mathrm{ln}\:{t}\:\rightarrow\:{u}'=\frac{\mathrm{1}}{{t}} \\ $$$$\:\:\:\:\:{v}'=\frac{\mathrm{4}{t}}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\:\rightarrow\:{v}=−\frac{\mathrm{2}}{{t}^{\mathrm{2}} −\mathrm{1}} \\ $$$$=−\frac{\mathrm{2ln}\:{t}}{{t}^{\mathrm{2}} −\mathrm{1}}+\mathrm{2}\int\frac{{dt}}{{t}\left({t}^{\mathrm{2}} −\mathrm{1}\right)}= \\ $$$$=−\frac{\mathrm{2ln}\:{t}}{{t}^{\mathrm{2}} −\mathrm{1}}+\mathrm{ln}\:\left({t}^{\mathrm{2}} −\mathrm{1}\right)\:−\mathrm{2ln}\:{t}\:= \\ $$$$=−\frac{\mathrm{2}{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} −\mathrm{1}}\mathrm{ln}\:{t}\:+\mathrm{ln}\:\left({t}^{\mathrm{2}} −\mathrm{1}\right)\:= \\ $$$$=−\frac{{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\mathrm{ln}\:\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)\:+\mathrm{ln}\:\left({x}^{\mathrm{2}} −\mathrm{1}+{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)\:+{C} \\ $$$$ \\ $$
Commented by john santu last updated on 12/May/20
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}\: \\ $$

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