dx-x-x-x-pleas-sir-help-me-

Question Number 83713 by mhmd last updated on 05/Mar/20

$$\int\frac{{dx}}{\:\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}}}\:\:\:{pleas}\:{sir}\:{help}\:{me} \\ $$

Commented by niroj last updated on 05/Mar/20

∫ (1/( (√(x+(√(x+(√x)))))))dx = ∫ (( 1)/((1/2)+((√(4x+1))/2)))dx= ∫ (2/(1+(√(4x+1))))dx put(√(4x+1)) =t 4x+1=t^2 4dx=2tdt dx= (t/2)dt ∫(2/(1+t)). (t/2)dt=∫(( t)/(t+1))dt = ∫(( t+1−1)/(t+1))dt= ∫dt−∫(( 1)/(t+1))dt = t −log (t+1)+C =(√(4x+1)) −log ((√(4x+1)) +1)+C//.

$$ \\ $$$$\:\:\int\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}}}\mathrm{dx} \\ $$$$\:=\:\int\:\frac{\:\:\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{4x}+\mathrm{1}}}{\mathrm{2}}}\mathrm{dx}=\:\int\:\frac{\mathrm{2}}{\mathrm{1}+\sqrt{\mathrm{4x}+\mathrm{1}}}\mathrm{dx} \\ $$$$\:\mathrm{put}\sqrt{\mathrm{4x}+\mathrm{1}}\:\:\:=\mathrm{t} \\ $$$$\:\:\:\:\:\:\:\mathrm{4x}+\mathrm{1}=\mathrm{t}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{4dx}=\mathrm{2tdt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{dx}=\:\frac{\mathrm{t}}{\mathrm{2}}\mathrm{dt} \\ $$$$\:\:\:\int\frac{\mathrm{2}}{\mathrm{1}+\mathrm{t}}.\:\frac{\mathrm{t}}{\mathrm{2}}\mathrm{dt}=\int\frac{\:\mathrm{t}}{\mathrm{t}+\mathrm{1}}\mathrm{dt} \\ $$$$=\:\int\frac{\:\:\mathrm{t}+\mathrm{1}−\mathrm{1}}{\mathrm{t}+\mathrm{1}}\mathrm{dt}=\:\int\mathrm{dt}−\int\frac{\:\mathrm{1}}{\mathrm{t}+\mathrm{1}}\mathrm{dt} \\ $$$$=\:\mathrm{t}\:−\mathrm{log}\:\left(\mathrm{t}+\mathrm{1}\right)+\mathrm{C} \\ $$$$\:=\sqrt{\mathrm{4x}+\mathrm{1}}\:\:\:\:−\mathrm{log}\:\left(\sqrt{\mathrm{4x}+\mathrm{1}}\:+\mathrm{1}\right)+\mathrm{C}//. \\ $$

Commented by MJS last updated on 05/Mar/20

how (√(x+(√(x+(√x)))))=(1/2)(1+(√(4x+1))) ?

$$\mathrm{how}\:\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\sqrt{\mathrm{4}{x}+\mathrm{1}}\right)\:? \\ $$

Commented by mhmd last updated on 05/Mar/20

$${im}\:{too}\:{dont}\:{know}\:{this}? \\ $$

Commented by jagoll last updated on 05/Mar/20

$$??? \\ $$

Commented by mhmd last updated on 05/Mar/20

$${how}\:{this}\:{sir}\:{can}\:{i}\:{help}\:{me}? \\ $$

Commented by MJS last updated on 05/Mar/20

if it′s (√(x+(√(x+(√(x+...)))))) to infinity: y=(√(x+(√(x+(√(x+...)))))) y^2 =x+y ⇒ y=(1/2)+(1/2)(√(4x+1)) but if it′s (√(x+(√(x+(√x))))) I′m afraid we cannot solve the integral

$$\mathrm{if}\:\mathrm{it}'\mathrm{s}\:\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+…}}}\:\mathrm{to}\:\mathrm{infinity}: \\ $$$${y}=\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+…}}} \\ $$$${y}^{\mathrm{2}} ={x}+{y} \\ $$$$\Rightarrow\:{y}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{4}{x}+\mathrm{1}} \\ $$$$\mathrm{but}\:\mathrm{if}\:\mathrm{it}'\mathrm{s}\:\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}}\:\mathrm{I}'\mathrm{m}\:\mathrm{afraid}\:\mathrm{we}\:\mathrm{cannot} \\ $$$$\mathrm{solve}\:\mathrm{the}\:\mathrm{integral} \\ $$

Commented by jagoll last updated on 05/Mar/20