1-1-x-dx-

Question Number 141578 by Khalmohmmad last updated on 20/May/21

$$\int\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}}}{dx}=? \\ $$

Answered by bramlexs22 last updated on 20/May/21

∫ ((√(1+x))/( (√x))) dx let (√(1+x)) = u ⇒x=u^2 −1 dx = 2u du I= ∫ (u/( (√(u^2 −1)))) 2u du I=∫ ((2u^2 )/( (√(u^2 −1)))) du u =sec t I=∫ ((2sec^2 t)/(tan t)) sec t tan t dt = 2∫ sec^3 t dt it easy to solve

$$\:\int\:\frac{\sqrt{\mathrm{1}+{x}}}{\:\sqrt{{x}}}\:{dx}\: \\ $$$$\:{let}\:\sqrt{\mathrm{1}+{x}}\:=\:{u}\:\Rightarrow{x}={u}^{\mathrm{2}} −\mathrm{1}\: \\ $$$${dx}\:=\:\mathrm{2}{u}\:{du}\: \\ $$$${I}=\:\int\:\frac{{u}}{\:\sqrt{{u}^{\mathrm{2}} −\mathrm{1}}}\:\mathrm{2}{u}\:{du}\: \\ $$$${I}=\int\:\frac{\mathrm{2}{u}^{\mathrm{2}} }{\:\sqrt{{u}^{\mathrm{2}} −\mathrm{1}}}\:{du} \\ $$$${u}\:=\mathrm{sec}\:{t}\: \\ $$$${I}=\int\:\frac{\mathrm{2sec}\:^{\mathrm{2}} {t}}{\mathrm{tan}\:{t}}\:\mathrm{sec}\:{t}\:\mathrm{tan}\:{t}\:{dt} \\ $$$$=\:\mathrm{2}\int\:\mathrm{sec}\:^{\mathrm{3}} {t}\:{dt}\: \\ $$$${it}\:{easy}\:{to}\:{solve} \\ $$

Answered by MJS_new last updated on 20/May/21

one possibility ∫(√(1+(1/x)))dx=∫(√((x+1)/x))dx= [t=(√x)+(√(x+1)) → dx=((2(√x)(√(x+1)))/( (√x)+(√(x+1))))dt] =(1/2)∫(((t^2 +1)^2 )/t^3 )dt=(1/2)∫tdt+∫(dt/t)+(1/2)∫(dt/t^3 )= =(t^2 /4)+ln t −(1/(4t^2 ))=(√x)(√(x+1))+ln ((√x)+(√(x+1))) +C

$$\mathrm{one}\:\mathrm{possibility} \\ $$$$\int\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}}}{dx}=\int\sqrt{\frac{{x}+\mathrm{1}}{{x}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{{x}}+\sqrt{{x}+\mathrm{1}}\:\rightarrow\:{dx}=\frac{\mathrm{2}\sqrt{{x}}\sqrt{{x}+\mathrm{1}}}{\:\sqrt{{x}}+\sqrt{{x}+\mathrm{1}}}{dt}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{{t}^{\mathrm{3}} }{dt}=\frac{\mathrm{1}}{\mathrm{2}}\int{tdt}+\int\frac{{dt}}{{t}}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{{t}^{\mathrm{3}} }= \\ $$$$=\frac{{t}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{ln}\:{t}\:−\frac{\mathrm{1}}{\mathrm{4}{t}^{\mathrm{2}} }=\sqrt{{x}}\sqrt{{x}+\mathrm{1}}+\mathrm{ln}\:\left(\sqrt{{x}}+\sqrt{{x}+\mathrm{1}}\right)\:+{C} \\ $$

Answered by MJS_new last updated on 20/May/21

another possibility ∫(√((x+1)/x))dx= [t=(√(x/(x+1))) → dx=2(√x)(√((x+1)^3 ))dt] =2∫(dt/((t^2 −1)^2 ))=−(t/(t^2 −1))−(1/2)ln ((t−1)/(t+1)) = =(√x)(√(x+1))+ln ((√x)+(√(x+1))) +C

$$\mathrm{another}\:\mathrm{possibility} \\ $$$$\int\sqrt{\frac{{x}+\mathrm{1}}{{x}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\frac{{x}}{{x}+\mathrm{1}}}\:\rightarrow\:{dx}=\mathrm{2}\sqrt{{x}}\sqrt{\left({x}+\mathrm{1}\right)^{\mathrm{3}} }{dt}\right] \\ $$$$=\mathrm{2}\int\frac{{dt}}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }=−\frac{{t}}{{t}^{\mathrm{2}} −\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\frac{{t}−\mathrm{1}}{{t}+\mathrm{1}}\:= \\ $$$$=\sqrt{{x}}\sqrt{{x}+\mathrm{1}}+\mathrm{ln}\:\left(\sqrt{{x}}+\sqrt{{x}+\mathrm{1}}\right)\:+{C} \\ $$

Answered by mathmax by abdo last updated on 20/May/21

I=∫(√(1+(1/x)))dx ⇒I=∫((√(1+x))/( (√x)))dx =_((√x)=t) ∫ ((√(1+t^2 ))/t)(2t)dt =2∫(√(1+t^2 ))dt =_(t=shy) 2∫ ch^2 y dy =∫(ch(2y)+1)dy =(1/2)sh(2y)+y +C =shy.chy +y +C =t(√(1+t^2 ))+argsh(t) +C =t(√(1+t^2 ))+log(t+(√(1+t^2 ))) +C I=(√x)(√(1+x))+log(x+(√(1+x))) +C

$$\mathrm{I}=\int\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}}\mathrm{dx}\:\Rightarrow\mathrm{I}=\int\frac{\sqrt{\mathrm{1}+\mathrm{x}}}{\:\sqrt{\mathrm{x}}}\mathrm{dx}\:=_{\sqrt{\mathrm{x}}=\mathrm{t}} \:\:\int\:\frac{\sqrt{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }}{\mathrm{t}}\left(\mathrm{2t}\right)\mathrm{dt} \\ $$$$=\mathrm{2}\int\sqrt{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:\:=_{\mathrm{t}=\mathrm{shy}} \:\:\mathrm{2}\int\:\mathrm{ch}^{\mathrm{2}} \mathrm{y}\:\mathrm{dy}\:=\int\left(\mathrm{ch}\left(\mathrm{2y}\right)+\mathrm{1}\right)\mathrm{dy} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sh}\left(\mathrm{2y}\right)+\mathrm{y}\:+\mathrm{C}\:=\mathrm{shy}.\mathrm{chy}\:+\mathrm{y}\:+\mathrm{C} \\ $$$$=\mathrm{t}\sqrt{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }+\mathrm{argsh}\left(\mathrm{t}\right)\:+\mathrm{C} \\ $$$$=\mathrm{t}\sqrt{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }+\mathrm{log}\left(\mathrm{t}+\sqrt{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\right)\:+\mathrm{C} \\ $$$$\mathrm{I}=\sqrt{\mathrm{x}}\sqrt{\mathrm{1}+\mathrm{x}}+\mathrm{log}\left(\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}}\right)\:+\mathrm{C} \\ $$

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