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Question Number 86431 by M±th+et£s last updated on 28/Mar/20
∫(√(x−(√(x^2 +1)) )) dx
$$\int\sqrt{{x}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:}\:{dx} \\ $$
Answered by jagoll last updated on 28/Mar/20
(√(x^2 +1)) = x−t x^2 +1 = x^2 −2tx+t^2 x = ((t^2 −1)/(2t)) ⇒ dx = ((4t^2 +2−2t^2 )/(4t^2 )) dt dx = ((2t^2 +2)/(4t^2 )) dt = ((t^2 +1)/(2t^2 )) dt ∫ (√( t)) ×((t^2 +1)/(2t^2 )) dt ∫ t^(1/2) ((1/2)+(1/2)t^(−2) ) dt = ∫ (1/2)t^(1/2) +(1/2)t^(−(3/2)) dt = (1/3)t(√(t )) − (1/( (√t))) + c = (1/3)(x−(√(x^2 +1)))^(3/2) −(1/( (√(x−(√(x^2 +1)))))) + c now it easy to solve
$$\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\:=\:\mathrm{x}−\mathrm{t}\: \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{1}\:=\:\mathrm{x}^{\mathrm{2}} −\mathrm{2tx}+\mathrm{t}^{\mathrm{2}} \\ $$$$\mathrm{x}\:=\:\frac{\mathrm{t}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2t}}\:\Rightarrow\:\mathrm{dx}\:=\:\frac{\mathrm{4t}^{\mathrm{2}} +\mathrm{2}−\mathrm{2t}^{\mathrm{2}} }{\mathrm{4t}^{\mathrm{2}} }\:\mathrm{dt} \\ $$$$\mathrm{dx}\:=\:\frac{\mathrm{2t}^{\mathrm{2}} +\mathrm{2}}{\mathrm{4t}^{\mathrm{2}} }\:\mathrm{dt}\:=\:\frac{\mathrm{t}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2t}^{\mathrm{2}} }\:\mathrm{dt} \\ $$$$\int\:\sqrt{\:\mathrm{t}}\:×\frac{\mathrm{t}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2t}^{\mathrm{2}} }\:\mathrm{dt}\: \\ $$$$\int\:\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{t}^{−\mathrm{2}} \right)\:\mathrm{dt} \\ $$$$=\:\int\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{2}}} +\frac{\mathrm{1}}{\mathrm{2}}\mathrm{t}^{−\frac{\mathrm{3}}{\mathrm{2}}} \:\mathrm{dt} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{3}}\mathrm{t}\sqrt{\mathrm{t}\:}\:−\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{t}}}\:+\:\mathrm{c} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{x}−\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}−\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}}}\:+\:\mathrm{c} \\ $$$$\mathrm{now}\:\mathrm{it}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve} \\ $$
Commented by M±th+et£s last updated on 28/Mar/20
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Commented by jagoll last updated on 28/Mar/20
why? i use Euler substitution sir
$$\mathrm{why}?\:\mathrm{i}\:\mathrm{use}\:\mathrm{Euler}\:\mathrm{substitution}\:\mathrm{sir} \\ $$
Commented by jagoll last updated on 28/Mar/20
t = x−(√(x^2 +1 )) sir (√(x^2 +1)) = x−t x^2 +1 = x^2 −2tx +t^2
$$\mathrm{t}\:=\:\mathrm{x}−\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}\:}\:\mathrm{sir} \\ $$$$\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\:=\:\mathrm{x}−\mathrm{t}\: \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{1}\:=\:\mathrm{x}^{\mathrm{2}} −\mathrm{2tx}\:+\mathrm{t}^{\mathrm{2}} \\ $$
Commented by jagoll last updated on 28/Mar/20
god bless you sir
$$\mathrm{god}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by MJS last updated on 28/Mar/20
(d/dx)[your solution]=−((x(√(x−(√(x^2 +1)))))/( (√(x^2 +1)))) ⇒ something went wrong
$$\frac{{d}}{{dx}}\left[\mathrm{your}\:\mathrm{solution}\right]=−\frac{{x}\sqrt{{x}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\Rightarrow\:\mathrm{something}\:\mathrm{went}\:\mathrm{wrong} \\ $$
Commented by MJS last updated on 28/Mar/20
we cannot solve t=(√(x−(√(x^2 +1)))) as if (√(x−(√(x^2 +1)))) was real if we square 2 times we get false solutions t^2 =x−(√(x^2 +1)) x−t^2 =(√(x^2 +1)) (x−t^2 )^2 =x^2 +1 −2t^2 x+t^4 −1=0 x=((t^4 −1)/(2t^2 )) but t=(√(x−(√(x^2 +1)))) x^2 +1=(((t^4 +1)^2 )/(4t^4 )) ⇒ (√(x^2 +1))=((t^4 +1)/(2t^2 )) x−(√(x^2 +1))=((t^4 −1)/(2t^2 ))−((t^4 +1)/(2t^2 ))=−(1/t^2 ) (√(x−(√(x^2 +1))))=i(1/t)≠t
$$\mathrm{we}\:\mathrm{cannot}\:\mathrm{solve} \\ $$$${t}=\sqrt{{x}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\mathrm{as}\:\mathrm{if}\:\sqrt{{x}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}\:\mathrm{was}\:\mathrm{real} \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{square}\:\mathrm{2}\:\mathrm{times}\:\mathrm{we}\:\mathrm{get}\:\mathrm{false}\:\mathrm{solutions} \\ $$$${t}^{\mathrm{2}} ={x}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$${x}−{t}^{\mathrm{2}} =\sqrt{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\left({x}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} ={x}^{\mathrm{2}} +\mathrm{1} \\ $$$$−\mathrm{2}{t}^{\mathrm{2}} {x}+{t}^{\mathrm{4}} −\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{{t}^{\mathrm{4}} −\mathrm{1}}{\mathrm{2}{t}^{\mathrm{2}} } \\ $$$$\mathrm{but} \\ $$$${t}=\sqrt{{x}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}} \\ $$$${x}^{\mathrm{2}} +\mathrm{1}=\frac{\left({t}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}{t}^{\mathrm{4}} }\:\Rightarrow\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}=\frac{{t}^{\mathrm{4}} +\mathrm{1}}{\mathrm{2}{t}^{\mathrm{2}} } \\ $$$${x}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}=\frac{{t}^{\mathrm{4}} −\mathrm{1}}{\mathrm{2}{t}^{\mathrm{2}} }−\frac{{t}^{\mathrm{4}} +\mathrm{1}}{\mathrm{2}{t}^{\mathrm{2}} }=−\frac{\mathrm{1}}{{t}^{\mathrm{2}} } \\ $$$$\sqrt{{x}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}=\mathrm{i}\frac{\mathrm{1}}{{t}}\neq{t} \\ $$
Commented by jagoll last updated on 28/Mar/20
(d/dx) [(1/3)( x−(√(x^2 +1 )))^(3/2) ]−(d/dx)[ (x−(√(x^2 +1)))^(−(1/2)) ] (1/2)(x−(√(x^2 +1)))^(1/2) (1−(x/( (√(x^2 +1))))) − (−(1/2))(x−(√(x^2 +1)))^(−(3/2)) (1−(x/( (√(x^2 +1)))))
$$\frac{\mathrm{d}}{\mathrm{dx}}\:\left[\frac{\mathrm{1}}{\mathrm{3}}\left(\:\mathrm{x}−\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}\:}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right]−\frac{\mathrm{d}}{\mathrm{dx}}\left[\:\left(\mathrm{x}−\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \right]\: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{x}−\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−\frac{\mathrm{x}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}}\right)\:− \\ $$$$\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\mathrm{x}−\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\right)^{−\frac{\mathrm{3}}{\mathrm{2}}} \left(\mathrm{1}−\frac{\mathrm{x}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}}\right) \\ $$
Commented by jagoll last updated on 28/Mar/20
my answer correct sir
$$\mathrm{my}\:\mathrm{answer}\:\mathrm{correct}\:\mathrm{sir} \\ $$
Commented by MJS last updated on 28/Mar/20
you changed it, now it′s correct
$$\mathrm{you}\:\mathrm{changed}\:\mathrm{it},\:{now}\:\mathrm{it}'\mathrm{s}\:\mathrm{correct} \\ $$
Answered by MJS last updated on 28/Mar/20
y=(√(x−(√(x^2 +1)))) ∉R but let me try... ∫(√(x−(√(x^2 +1))))dx=i∫(√(−x+(√(x^2 +1))))dx= [t=(√(−x+(√(x^2 +1)))) → dx=−((2(√(x^2 +1)))/( (√(−x+(√(x^2 +1))))))] =−i∫(t^2 −(1/t^2 ))dt=−i((t^3 /3)−(1/t))= =((2i)/3)(2x+(√(x^2 +1)))(√(−x+(√(x^2 +1))))= =(2/3)(2x+(√(x^2 +1)))(√(x−(√(x^2 +1)))) +C
$${y}=\sqrt{{x}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}\:\notin\mathbb{R} \\ $$$$\mathrm{but}\:\mathrm{let}\:\mathrm{me}\:\mathrm{try}… \\ $$$$\int\sqrt{{x}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{dx}=\mathrm{i}\int\sqrt{−{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{−{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}\:\rightarrow\:{dx}=−\frac{\mathrm{2}\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{\:\sqrt{−{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}}\right] \\ $$$$=−\mathrm{i}\int\left({t}^{\mathrm{2}} −\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right){dt}=−\mathrm{i}\left(\frac{{t}^{\mathrm{3}} }{\mathrm{3}}−\frac{\mathrm{1}}{{t}}\right)= \\ $$$$=\frac{\mathrm{2i}}{\mathrm{3}}\left(\mathrm{2}{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)\sqrt{−{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}= \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{2}{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)\sqrt{{x}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}\:+{C} \\ $$

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