Infinite Radicals!

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

$\dfrac{3}{4} \leq x \leq 2$ .

$\implies \displaystyle\int_{\frac{3}{4}}^{2} (\sqrt{x + \sqrt{x + \sqrt{x + ... }}}) \:\ dx = \dfrac{1}{2}\displaystyle\int_{\frac{3}{4}}^{2} (1 + \sqrt{1 + 4x}) \:\ dx =$

$\dfrac{1}{2}(x + \dfrac{1}{6}(1 + 4x)^{\dfrac{3}{2}})|_{\frac{3}{4}}^{2} =$

$\dfrac{1}{2}(2 + \dfrac{19}{6} - \dfrac{3}{4}) = \dfrac{53}{24} = \dfrac{a}{b} \implies a + b =\boxed{77}$ .

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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.