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Question Number 82566 by jagoll last updated on 22/Feb/20
∫ (√((x+1)/x)) dx = ?
$$\int\:\sqrt{\frac{{x}+\mathrm{1}}{{x}}}\:{dx}\:=\:? \\ $$
Commented by mathmax by abdo last updated on 22/Feb/20
let I=∫(√((x+1)/x))dx changement (√((x+1)/x))=t give ((x+1)/x)=t^2 ⇒ x+1 =xt^2 ⇒(1−t^2 )x=−1 ⇒x =(1/(t^2 −1)) ⇒dx =((−2tdt)/((t^2 −1)^2 )) ⇒ I =−2 ∫ t^2 ×(dt/((t^2 −1)^2 )) =∫ ((((−2t))/((t^2 −1)^2 )))tdt by parts u^′ =((−2t)/((t^2 −1)^2 )) and v=t ⇒ I =(t/(t^2 −1))−∫ (1/(t^2 −1))dt =(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+1)/x))/(((x+1)/x)−1))−(1/2)ln∣(((√((x+1)/x))−1)/( (√((x+1)/x))+1))∣ +C =x(√((x+1)/x)) −(1/2)ln∣(((√((x+1)/x))−1)/( (√((x+1)/x))+1))∣ +C.
$${let}\:{I}=\int\sqrt{\frac{{x}+\mathrm{1}}{{x}}}{dx}\:{changement}\:\sqrt{\frac{{x}+\mathrm{1}}{{x}}}={t}\:{give}\:\frac{{x}+\mathrm{1}}{{x}}={t}^{\mathrm{2}} \:\Rightarrow \\ $$$${x}+\mathrm{1}\:={xt}^{\mathrm{2}} \:\Rightarrow\left(\mathrm{1}−{t}^{\mathrm{2}} \right){x}=−\mathrm{1}\:\Rightarrow{x}\:=\frac{\mathrm{1}}{{t}^{\mathrm{2}} −\mathrm{1}}\:\Rightarrow{dx}\:=\frac{−\mathrm{2}{tdt}}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${I}\:=−\mathrm{2}\:\int\:{t}^{\mathrm{2}} ×\frac{{dt}}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\:\:=\int\:\left(\frac{\left(−\mathrm{2}{t}\right)}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\right){tdt}\:{by}\:{parts}\:{u}^{'} \:=\frac{−\mathrm{2}{t}}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${and}\:{v}={t}\:\Rightarrow\:{I}\:=\frac{{t}}{{t}^{\mathrm{2}} −\mathrm{1}}−\int\:\frac{\mathrm{1}}{{t}^{\mathrm{2}} −\mathrm{1}}{dt} \\ $$$$=\frac{{t}}{{t}^{\mathrm{2}} −\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\int\left(\frac{\mathrm{1}}{{t}−\mathrm{1}}−\frac{\mathrm{1}}{{t}+\mathrm{1}}\right){dt} \\ $$$$=\frac{{t}}{{t}^{\mathrm{2}} −\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{{t}−\mathrm{1}}{{t}+\mathrm{1}}\mid\:+{c}\:=\frac{\sqrt{\frac{{x}+\mathrm{1}}{{x}}}}{\frac{{x}+\mathrm{1}}{{x}}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{\sqrt{\frac{{x}+\mathrm{1}}{{x}}}−\mathrm{1}}{\:\sqrt{\frac{{x}+\mathrm{1}}{{x}}}+\mathrm{1}}\mid\:+{C} \\ $$$$={x}\sqrt{\frac{{x}+\mathrm{1}}{{x}}}\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{\sqrt{\frac{{x}+\mathrm{1}}{{x}}}−\mathrm{1}}{\:\sqrt{\frac{{x}+\mathrm{1}}{{x}}}+\mathrm{1}}\mid\:+{C}. \\ $$$$ \\ $$
Answered by jagoll last updated on 22/Feb/20
Commented by mathmax by abdo last updated on 22/Feb/20
sir jagol (√((x+1)/x))≠((√(x+1))/( (√x)))
$${sir}\:{jagol}\:\sqrt{\frac{{x}+\mathrm{1}}{{x}}}\neq\frac{\sqrt{{x}+\mathrm{1}}}{\:\sqrt{{x}}} \\ $$
Commented by john santu last updated on 22/Feb/20
why (√((x+1)/x)) ≠ ((√(x+1))/( (√x))) (√(2/3)) ≠ ((√2)/( (√3))) ?? i think it same sir
$${why}\:\sqrt{\frac{{x}+\mathrm{1}}{{x}}}\:\neq\:\frac{\sqrt{{x}+\mathrm{1}}}{\:\sqrt{{x}}}\: \\ $$$$\sqrt{\frac{\mathrm{2}}{\mathrm{3}}}\:\neq\:\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}}\:??\: \\ $$$${i}\:{think}\:{it}\:{same}\:{sir}\: \\ $$
Commented by mathmax by abdo last updated on 23/Feb/20
no sir (√((x+1)/x))is defined on]−∞,−1[∪]0,+∞[ but ((√(x+1))/( (√x))) is defined on ]0,+∞[ (its not equals)
$$\left.{no}\:{sir}\:\sqrt{\frac{{x}+\mathrm{1}}{{x}}}{is}\:{defined}\:{on}\right]−\infty,−\mathrm{1}\left[\cup\right]\mathrm{0},+\infty\left[\:\:{but}\:\frac{\sqrt{{x}+\mathrm{1}}}{\:\sqrt{{x}}}\right. \\ $$$$\left.{is}\:{defined}\:{on}\:\right]\mathrm{0},+\infty\left[\:\:\left({its}\:{not}\:{equals}\right)\right. \\ $$
Answered by niroj last updated on 22/Feb/20
Let I =∫(√(((x+1)^2 )/(x(x+1)))) dx = ∫ ((x+1)/( (√(x(x+1)))))dx = ∫ (x/( (√(x^2 +x))))dx+ ∫(( 1)/( (√(x^2 +x))))dx I_1 =∫ (x/( (√(x^2 +x))))dx & I_2 =∫ (1/( (√(x^2 +x))))dx For(I_1 ),put x^2 =t⇒ x=(√t) 2xdx=dt xdx=(1/2)dt I_1 = ∫ (1/(2(√(t+(√t)))))dt I_1 = (1/2)∫ (( 1)/( (√(((√t) )^2 +2.(√t)(1/2)+(1/4)−(1/4)))))dt I=I_1 +I_2 = (1/2)∫(1/( (√(((√t) +(1/2))^2 −((1/2))^2 ))))dt +∫ (1/( (√((x)^2 +2.x.(1/2)+(1/4)−(1/4)))))dx = (1/2)log( (√t)+(1/2)+ (√(((√t) +(1/(2 )))^2 −(1/4))) )+∫(1/( (√((x+(1/2))^2 −(1/4)))))dx = (1/2)log((√t)+(1/2)+(√(t+(√t))) )+log(x+(1/2)+(√(x^2 +x)) )+C = (1/2)log(x+(1/2)+(√(x^2 +x)) )+log(x+(1/2)+(√(x^2 +x)) )+C = log(x+(1/2)+(√(x^2 +x)) )^(1/2) .(x+(1/2)+(√(x^2 +x)) )+C = log(x+(1/2)+(√(x^2 +x)) )^(3/2) +C = (3/2)log(((2x+1+2(√(x^2 +x)) )/2))+C = (3/2)log(((2x+1+2(√(x(x+1))))/2))+c = (3/2)log ((((√x) + (√(x+1)))^2 )/2)+c = (3/2)[ log ((√x) +(√(x+1)) )^2 −log2]+c = (3/2).2log((√x) +(√(x+1)) )−(3/2)log2+C = log ((√x) +(√(x+1)) )^3 − (3/2)log2+C//.
$$\:\mathrm{Let}\:\mathrm{I}\:=\int\sqrt{\frac{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{x}\left(\mathrm{x}+\mathrm{1}\right)}}\:\:\:\mathrm{dx} \\ $$$$=\:\int\:\frac{\mathrm{x}+\mathrm{1}}{\:\sqrt{\mathrm{x}\left(\mathrm{x}+\mathrm{1}\right)}}\mathrm{dx} \\ $$$$\:=\:\int\:\frac{\mathrm{x}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}}}\mathrm{dx}+\:\int\frac{\:\mathrm{1}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}}}\mathrm{dx} \\ $$$$\:\:\mathrm{I}_{\mathrm{1}} =\int\:\frac{\mathrm{x}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}}}\mathrm{dx}\:\&\:\mathrm{I}_{\mathrm{2}} =\int\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}}}\mathrm{dx} \\ $$$$\:\:\mathrm{For}\left(\mathrm{I}_{\mathrm{1}} \right),\mathrm{put}\:\mathrm{x}^{\mathrm{2}} =\mathrm{t}\Rightarrow\:\mathrm{x}=\sqrt{\mathrm{t}} \\ $$$$\:\:\:\:\:\:\mathrm{2xdx}=\mathrm{dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{xdx}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{dt} \\ $$$$\:\:\:\mathrm{I}_{\mathrm{1}} =\:\int\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{t}+\sqrt{\mathrm{t}}}}\mathrm{dt}\: \\ $$$$\mathrm{I}_{\mathrm{1}} =\:\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\:\mathrm{1}}{\:\sqrt{\left(\sqrt{\mathrm{t}}\:\right)^{\mathrm{2}} +\mathrm{2}.\sqrt{\mathrm{t}}\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}}}\mathrm{dt} \\ $$$$\:\mathrm{I}=\mathrm{I}_{\mathrm{1}} +\mathrm{I}_{\mathrm{2}} \\ $$$$\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{\:\sqrt{\left(\sqrt{\mathrm{t}}\:\:+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }}\mathrm{dt}\:+\int\:\frac{\mathrm{1}}{\:\sqrt{\left(\mathrm{x}\right)^{\mathrm{2}} +\mathrm{2}.\mathrm{x}.\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}}}\mathrm{dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\left(\:\sqrt{\mathrm{t}}+\frac{\mathrm{1}}{\mathrm{2}}+\:\sqrt{\left(\sqrt{\mathrm{t}}\:\:+\frac{\mathrm{1}}{\mathrm{2}\:}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}}\:\right)+\int\frac{\mathrm{1}}{\:\sqrt{\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}}}\mathrm{dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\left(\sqrt{\mathrm{t}}+\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\mathrm{t}+\sqrt{\mathrm{t}}}\:\:\right)+\mathrm{log}\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}}\:\right)+\mathrm{C} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}}\:\right)+\mathrm{log}\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}}\:\right)+\mathrm{C} \\ $$$$=\:\mathrm{log}\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}}\:\right)^{\frac{\mathrm{1}}{\mathrm{2}}} .\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}}\:\right)+\mathrm{C} \\ $$$$\:=\:\mathrm{log}\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}}\:\:\right)^{\frac{\mathrm{3}}{\mathrm{2}}} +\mathrm{C} \\ $$$$\:=\:\frac{\mathrm{3}}{\mathrm{2}}\mathrm{log}\left(\frac{\mathrm{2x}+\mathrm{1}+\mathrm{2}\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}}\:}{\mathrm{2}}\right)+\mathrm{C} \\ $$$$\:=\:\frac{\mathrm{3}}{\mathrm{2}}\mathrm{log}\left(\frac{\mathrm{2x}+\mathrm{1}+\mathrm{2}\sqrt{\mathrm{x}\left(\mathrm{x}+\mathrm{1}\right)}}{\mathrm{2}}\right)+\mathrm{c} \\ $$$$=\:\frac{\mathrm{3}}{\mathrm{2}}\mathrm{log}\:\:\frac{\left(\sqrt{\mathrm{x}}\:\:+\:\sqrt{\mathrm{x}+\mathrm{1}}\right)^{\mathrm{2}} }{\mathrm{2}}+\mathrm{c} \\ $$$$=\:\frac{\mathrm{3}}{\mathrm{2}}\left[\:\mathrm{log}\:\left(\sqrt{\mathrm{x}}\:\:+\sqrt{\mathrm{x}+\mathrm{1}}\:\right)^{\mathrm{2}} −\mathrm{log2}\right]+\mathrm{c} \\ $$$$=\:\frac{\mathrm{3}}{\mathrm{2}}.\mathrm{2log}\left(\sqrt{\mathrm{x}}\:\:+\sqrt{\mathrm{x}+\mathrm{1}}\:\right)−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{log2}+\mathrm{C} \\ $$$$=\:\:\mathrm{log}\:\left(\sqrt{\mathrm{x}}\:\:\:+\sqrt{\mathrm{x}+\mathrm{1}}\:\:\right)^{\mathrm{3}} −\:\frac{\mathrm{3}}{\mathrm{2}}\mathrm{log2}+\mathrm{C}//. \\ $$$$ \\ $$$$ \\ $$

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