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

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

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

Answered by Aziztisffola last updated on 11/Jul/20