Infinite Radicals!!

Let $u= \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{\cdots}}}}$ . Then

$\begin{aligned} u & = \sqrt{x+u} \\ u^2 & = x + u \\ u^2 - u & = x \\ \implies (2u-1) \ du & = dx \end{aligned}$

And when $\begin{cases} x = 6 & \implies u^2 - u - 6 = (u-3)(u+2) = 0 & \implies u = 3 \text{ since }u> 0 \\ x = 2 & \implies u^2 - u - 2 = (u-2)(u+1) = 0 & \implies u = 2 \text{ since }u> 0 \end{cases}$

Now we have

$\begin{aligned} I & = \int_2^6 \frac 1{\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{\cdots}}}}} dx \\ & = \int_2^3 \frac {2u-1}u du \\ & = \int_2^3 \left(2 - \frac 1u\right) du \\ & = 2u - \ln u \ \big|_2^3 \\ & = 2 + \ln \left(\frac 23\right) \end{aligned}$

Therefore $a+b = 2+3 = \boxed 5$ .

Let $y = \sqrt{x + \sqrt{x + \sqrt{x + ... }}} \implies y^2 - x = \sqrt{x + \sqrt{x + \sqrt{x + ... }}} = y$

$\implies y^2 - y - x = 0 \implies y = \dfrac{1 + \sqrt{1 + 4x}}{2}$ dropping the negative root for

$2 \leq x \leq 6$ .

Let $w(x) = \dfrac{1}{y(x)}$

$\implies \displaystyle\int_{2}^{6} w(x) dx =$ $2\displaystyle\int_{2}^{6} (\dfrac{dx}{1 + \sqrt{4x + 1}})$

Let $u^2 = 4x + 1 \implies dx = \dfrac{1}{2}u du \implies 2\displaystyle\int_{2}^{6} (\dfrac{dx}{1 + \sqrt{4x + 1}}) =$

$\displaystyle\int_{3}^{5} \dfrac{u}{u + 1} \:\ du = \displaystyle\int_{3}^{5} (1 - \dfrac{1}{u + 1}) \:\ du =$

$u - \ln(u + 1)|_{3}^{5} = 2 + \ln(\dfrac{2}{3}) = a + \ln(\dfrac{a}{b}) \implies a + b = \boxed{5}$ .

Note:

Letting $z_{1} = \sqrt{x}$ and $z_{n + 1} = \sqrt{x + z_{n}}$ and assuming $\lim_{n \rightarrow \infty} z_{n}$ exists

$\implies$

$y = \lim_{n \rightarrow \infty} z_{n} = \lim_{n \rightarrow \infty} z_{n + 1} = \lim_{n \rightarrow \infty} \sqrt{x + z_{n}}$

and $\lim_{n \rightarrow \infty} S_{n} = L \implies \lim_{n \rightarrow \infty} (S_{n})^2 = L^2$

$\implies y^2 = \lim_{n \rightarrow \infty} x + z_{n} = x + \lim_{n \rightarrow \infty} z_{n} = x + y \implies$

$\implies y^2 - y - x = 0 \implies y = \dfrac{1 + \sqrt{1 + 4x}}{2} \:\$ same as above from here.