Question Number 200980 by sonukgindia last updated on 27/Nov/23
Answered by MM42 last updated on 27/Nov/23
![(√(x+(√(x+(√(x...))))))=y⇒ x=y^2 −y⇒dx=(2y−1)dy y=((1+(√(1+4x)))/2) ⇒ x=0→y=1 ; x=1→y=((1+(√5))/2) ⇒∫_1 ^((1+(√5))/2) ((2y−1)/y) dy =(2y−lny)]_1 ^((1+(√5))/2) =(1+(√5)−ln(((1+(√5))/2) )−2) =(√5)−1−ln(((1+(√5))/2)) ✓](https://www.tinkutara.com/question/Q200988.png)
$$\sqrt{{x}+\sqrt{{x}+\sqrt{{x}…}}}={y}\Rightarrow\:{x}={y}^{\mathrm{2}} −{y}\Rightarrow{dx}=\left(\mathrm{2}{y}−\mathrm{1}\right){dy} \\ $$$${y}=\frac{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4}{x}}}{\mathrm{2}}\:\:\:\:\:\Rightarrow\:\:\:{x}=\mathrm{0}\rightarrow{y}=\mathrm{1}\:\:\:\:;\:\:{x}=\mathrm{1}\rightarrow{y}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\left.\Rightarrow\int_{\mathrm{1}} ^{\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}} \:\:\frac{\mathrm{2}{y}−\mathrm{1}}{{y}}\:{dy}\:=\left(\mathrm{2}{y}−{lny}\right)\right]_{\mathrm{1}} ^{\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}} \:\: \\ $$$$=\left(\mathrm{1}+\sqrt{\mathrm{5}}−{ln}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\right)−\mathrm{2}\right) \\ $$$$=\sqrt{\mathrm{5}}−\mathrm{1}−{ln}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\:\:\checkmark \\ $$$$ \\ $$