Integration and radicals (2)

$\begin{aligned} y & = \sqrt{x+x\sqrt{x+x\sqrt{x+x\sqrt{\cdots}}}} \\ & = \sqrt{x+xy} \\ y^2 & = x+xy \\ x & = \frac {y^2}{1+y} \\ dx & = \frac {2y(y+1)-y^2}{(y+1)^2} dy \\ & = \frac {y^2+2y}{(y+1)^2}dy \\ & = \left(1 - \frac 1{(y+1)^2}\right) dy \end{aligned}$

Therefore, we have:

$\begin{aligned} I & = \int_0^1 \sqrt{x+x\sqrt{x+x\sqrt{x+x\sqrt{\cdots}}}} dx \\ & = \int_0^1 y \ dx \\ & = \int_0^\phi \left(y - \frac y{(y+1)^2}\right) dy & \small \color{#3D99F6} \text{Note: }\begin{cases} x = 1 & \implies y = \phi = \frac {\sqrt 5 + 1}2 \\ x = 0 & \implies y = 0 \end{cases} \\ & = \int_0^\phi \left(y - \frac {2y+2-2}{2(y+1)^2}\right) dy \\ & = \int_0^\phi \left(y - \frac {2y+2}{2(y+1)^2} + \frac 1{(y+1)^2} \right) dy \\ & = \left[\frac {y^2}2 - \ln (y+1) - \frac 1{y+1} \right]_0^\phi \\ & = \frac {\phi^2}2 - \ln (\phi+1) - \frac 1{\phi+1} + 1 \\ & = \frac {3+\sqrt 5}4 - \ln \left(\frac {3+\sqrt 5}2\right) - \frac 2{3+\sqrt 5} + 1 \\ & = \frac {3\sqrt 5+1}4 - \ln \left(\frac {3+\sqrt 5}2\right) \end{aligned}$

$\implies a + b = 3+5 = \boxed{8}$