Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 102894 by bramlex last updated on 11/Jul/20

∫ (√(x+(√(x+(√(x+(√(x+(√(x+...)))))))))) dx

$$\int\:\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+...}}}}}\:\mathrm{dx}\: \\ $$

Answered by bemath last updated on 11/Jul/20

y=(√(x+(√(x+(√(x+(√(x+(√(...)))))))))) y^2 = x+y ⇒y^2 −y−x = 0 y = ((1 + (√(1+4x)))/2) I=∫ (((1+(√(1+4x)))/2)) dx I = (x/2) +(1/2)∫ (1+4x)^(1/2) dx I= (x/2)+ (1/2). (2/3).(1/4)(√((1+4x)^3 )) + C I= (x/2) + (1/(12)) (√((1+4x)^3 )) + C

$${y}=\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+\sqrt{...}}}}}\: \\ $$$${y}^{\mathrm{2}} \:=\:{x}+{y}\:\Rightarrow{y}^{\mathrm{2}} −{y}−{x}\:=\:\mathrm{0} \\ $$$${y}\:=\:\frac{\mathrm{1}\:+\:\sqrt{\mathrm{1}+\mathrm{4}{x}}}{\mathrm{2}} \\ $$$${I}=\int\:\left(\frac{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4}{x}}}{\mathrm{2}}\right)\:{dx}\: \\ $$$${I}\:=\:\frac{{x}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}\int\:\left(\mathrm{1}+\mathrm{4}{x}\right)^{\mathrm{1}/\mathrm{2}} \:{dx}\: \\ $$$${I}=\:\frac{{x}}{\mathrm{2}}+\:\frac{\mathrm{1}}{\mathrm{2}}.\:\frac{\mathrm{2}}{\mathrm{3}}.\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\left(\mathrm{1}+\mathrm{4}{x}\right)^{\mathrm{3}} }\:+\:{C}\: \\ $$$${I}=\:\frac{{x}}{\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{12}}\:\sqrt{\left(\mathrm{1}+\mathrm{4}{x}\right)^{\mathrm{3}} }\:+\:{C}\: \\ $$

Answered by Dwaipayan Shikari last updated on 11/Jul/20

(√(x+(√(x+(√(x..))))))=y x+y=y^2 y^2 −y−x=0 y=((1+(√(1+4x)))/2) ∫((1+(√(1+4x)))/2)=(x/2)+(1/8)∫4(√(1+4x))dx=(x/2)+(1/4)∫u^2 du (take (√(1+4x))=u =(x/2)+(1/(12))u^3 +C=(x/2)+(1/(12))(1+4x)^(3/2) +Constant If it is ∫(√(x(√(x(√(x(√(x(√(x....∞))))))))))dx=∫y=∫xdx=(x^2 /2)+C y^2 =xy y^2 −xy=0 y=0 or y=x

$$\sqrt{{x}+\sqrt{{x}+\sqrt{{x}..}}}={y} \\ $$$${x}+{y}={y}^{\mathrm{2}} \\ $$$${y}^{\mathrm{2}} −{y}−{x}=\mathrm{0} \\ $$$${y}=\frac{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4}{x}}}{\mathrm{2}} \\ $$$$\int\frac{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4}{x}}}{\mathrm{2}}=\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{8}}\int\mathrm{4}\sqrt{\mathrm{1}+\mathrm{4}{x}}{dx}=\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}\int{u}^{\mathrm{2}} {du}\:\:\left({take}\:\sqrt{\mathrm{1}+\mathrm{4}{x}}={u}\right. \\ $$$$=\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{12}}{u}^{\mathrm{3}} +{C}=\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{12}}\left(\mathrm{1}+\mathrm{4}{x}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} +{Constant} \\ $$$${If}\:{it}\:{is} \\ $$$$\int\sqrt{{x}\sqrt{{x}\sqrt{{x}\sqrt{{x}\sqrt{{x}....\infty}}}}}{dx}=\int{y}=\int{xdx}=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{C} \\ $$$${y}^{\mathrm{2}} ={xy} \\ $$$${y}^{\mathrm{2}} −{xy}=\mathrm{0}\:\:{y}=\mathrm{0}\:\:{or}\:{y}={x} \\ $$$$ \\ $$

Answered by Aziztisffola last updated on 11/Jul/20

Terms of Service

Privacy Policy

Contact: info@tinkutara.com