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Question Number 136840 by physicstutes last updated on 26/Mar/21
 ∫(√(x/(x−1))) dx
$$\:\int\sqrt{\frac{{x}}{{x}−\mathrm{1}}}\:{dx} \\ $$
Answered by Olaf last updated on 26/Mar/21
F(x) = ∫(√(x/(x−1))) dx  F(u) =^(x=u^2 )  ∫(√(u^2 /(u^2 −1))) 2udu  F(u) = 2∫(u^2 /( (√(u^2 −1)))) du  F(u) = 2∫((√(u^2 −1))+(1/( (√(u^2 −1))))) du  F(u) = 2u(√(u^2 −1))−2∫u((2u)/(2(√(u^2 −1)))) du+2argchu  F(u) = 2argchu+2u(√(u^2 −1))−F(u)  F(u) = argchu+u(√(u^2 −1))  F(x) = argch(√x)+(√x).(√(x−1))
$$\mathrm{F}\left({x}\right)\:=\:\int\sqrt{\frac{{x}}{{x}−\mathrm{1}}}\:{dx} \\ $$$$\mathrm{F}\left({u}\right)\:\overset{{x}={u}^{\mathrm{2}} } {=}\:\int\sqrt{\frac{{u}^{\mathrm{2}} }{{u}^{\mathrm{2}} −\mathrm{1}}}\:\mathrm{2}{udu} \\ $$$$\mathrm{F}\left({u}\right)\:=\:\mathrm{2}\int\frac{{u}^{\mathrm{2}} }{\:\sqrt{{u}^{\mathrm{2}} −\mathrm{1}}}\:{du} \\ $$$$\mathrm{F}\left({u}\right)\:=\:\mathrm{2}\int\left(\sqrt{{u}^{\mathrm{2}} −\mathrm{1}}+\frac{\mathrm{1}}{\:\sqrt{{u}^{\mathrm{2}} −\mathrm{1}}}\right)\:{du} \\ $$$$\mathrm{F}\left({u}\right)\:=\:\mathrm{2}{u}\sqrt{{u}^{\mathrm{2}} −\mathrm{1}}−\mathrm{2}\int{u}\frac{\mathrm{2}{u}}{\mathrm{2}\sqrt{{u}^{\mathrm{2}} −\mathrm{1}}}\:{du}+\mathrm{2argch}{u} \\ $$$$\mathrm{F}\left({u}\right)\:=\:\mathrm{2argch}{u}+\mathrm{2}{u}\sqrt{{u}^{\mathrm{2}} −\mathrm{1}}−\mathrm{F}\left({u}\right) \\ $$$$\mathrm{F}\left({u}\right)\:=\:\mathrm{argch}{u}+{u}\sqrt{{u}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\mathrm{argch}\sqrt{{x}}+\sqrt{{x}}.\sqrt{{x}−\mathrm{1}} \\ $$
Commented by physicstutes last updated on 26/Mar/21
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by MJS_new last updated on 26/Mar/21
∫(√(x/(x−1)))dx=       [t=(√x)+(√(x−1)) → dx=((2(√x)(√(x−1)))/( (√x)+(√(x−1))))]  =∫((t/2)+(1/t)+(1/(2t^3 )))dt=(t^2 /4)+ln t −(1/(4t^2 ))=  =ln ((√x)+(√(x−1))) +(√x)(√(x−1))+C
$$\int\sqrt{\frac{{x}}{{x}−\mathrm{1}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{{x}}+\sqrt{{x}−\mathrm{1}}\:\rightarrow\:{dx}=\frac{\mathrm{2}\sqrt{{x}}\sqrt{{x}−\mathrm{1}}}{\:\sqrt{{x}}+\sqrt{{x}−\mathrm{1}}}\right] \\ $$$$=\int\left(\frac{{t}}{\mathrm{2}}+\frac{\mathrm{1}}{{t}}+\frac{\mathrm{1}}{\mathrm{2}{t}^{\mathrm{3}} }\right){dt}=\frac{{t}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{ln}\:{t}\:−\frac{\mathrm{1}}{\mathrm{4}{t}^{\mathrm{2}} }= \\ $$$$=\mathrm{ln}\:\left(\sqrt{{x}}+\sqrt{{x}−\mathrm{1}}\right)\:+\sqrt{{x}}\sqrt{{x}−\mathrm{1}}+{C} \\ $$
Answered by mathmax by abdo last updated on 27/Mar/21
I=∫ (√(x/(x−1)))dx we do the changement (x/(x−1))=t^2  ⇒  x=t^2 x−t^2  ⇒(1−t^2 )x=−t^2  ⇒x=((−t^2 )/(1−t^2 ))=(t^2 /(t^2 −1)) ⇒  (dx/dt)=((2t(t^2 −1)−t^2 (2t))/((t^2 −1)^2 ))=((−2t)/((t^2 −1)^2 )) ⇒I=∫ t(((−2t)/((t^2 −1)^2 )))dt  by parts u=t and v^′  =((−2t)/((t^2 −1)^2 ))=−2t(t^2 −1)^(−2)  ⇒  I =(t/(t^2 −1))−∫ (dt/(t^2 −1)) =(t/(t^2 −1))−(1/2)∫((1/(t−1))−(1/(t+1)))dt  =(t/(t^2 −1))−(1/2)ln∣((t−1)/(t+1))∣ +C  =((√(x/(x−1)))/((x/(x−1))−1))−(1/2)ln∣(((√(x/(x−1)))−1)/( (√(x/(x−1)))+1))∣ +C
$$\mathrm{I}=\int\:\sqrt{\frac{\mathrm{x}}{\mathrm{x}−\mathrm{1}}}\mathrm{dx}\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\frac{\mathrm{x}}{\mathrm{x}−\mathrm{1}}=\mathrm{t}^{\mathrm{2}} \:\Rightarrow \\ $$$$\mathrm{x}=\mathrm{t}^{\mathrm{2}} \mathrm{x}−\mathrm{t}^{\mathrm{2}} \:\Rightarrow\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)\mathrm{x}=−\mathrm{t}^{\mathrm{2}} \:\Rightarrow\mathrm{x}=\frac{−\mathrm{t}^{\mathrm{2}} }{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }=\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{t}^{\mathrm{2}} −\mathrm{1}}\:\Rightarrow \\ $$$$\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\mathrm{2t}\left(\mathrm{t}^{\mathrm{2}} −\mathrm{1}\right)−\mathrm{t}^{\mathrm{2}} \left(\mathrm{2t}\right)}{\left(\mathrm{t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }=\frac{−\mathrm{2t}}{\left(\mathrm{t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\mathrm{I}=\int\:\mathrm{t}\left(\frac{−\mathrm{2t}}{\left(\mathrm{t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\right)\mathrm{dt} \\ $$$$\mathrm{by}\:\mathrm{parts}\:\mathrm{u}=\mathrm{t}\:\mathrm{and}\:\mathrm{v}^{'} \:=\frac{−\mathrm{2t}}{\left(\mathrm{t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }=−\mathrm{2t}\left(\mathrm{t}^{\mathrm{2}} −\mathrm{1}\right)^{−\mathrm{2}} \:\Rightarrow \\ $$$$\mathrm{I}\:=\frac{\mathrm{t}}{\mathrm{t}^{\mathrm{2}} −\mathrm{1}}−\int\:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} −\mathrm{1}}\:=\frac{\mathrm{t}}{\mathrm{t}^{\mathrm{2}} −\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\int\left(\frac{\mathrm{1}}{\mathrm{t}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{t}+\mathrm{1}}\right)\mathrm{dt} \\ $$$$=\frac{\mathrm{t}}{\mathrm{t}^{\mathrm{2}} −\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\frac{\mathrm{t}−\mathrm{1}}{\mathrm{t}+\mathrm{1}}\mid\:+\mathrm{C} \\ $$$$=\frac{\sqrt{\frac{\mathrm{x}}{\mathrm{x}−\mathrm{1}}}}{\frac{\mathrm{x}}{\mathrm{x}−\mathrm{1}}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\frac{\sqrt{\frac{\mathrm{x}}{\mathrm{x}−\mathrm{1}}}−\mathrm{1}}{\:\sqrt{\frac{\mathrm{x}}{\mathrm{x}−\mathrm{1}}}+\mathrm{1}}\mid\:+\mathrm{C} \\ $$

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