Infinite Radicals!

$x + 2 = \sqrt{(x + 2)^2} = \sqrt{1 + (x^2 + 4x + 3)} = \sqrt{1 + (x + 1)(x + 3)} =$

$\sqrt{1 + (x + 1)\sqrt{(x + 3)^2}} = \sqrt{1 + (x + 1)\sqrt{1 + x^2 + 6x + 8}} =$

$\sqrt{1 + (x + 1)\sqrt{1 + (x + 2)(x + 4)}} = \sqrt{1 + (x + 1)\sqrt{1 + (x + 2)\sqrt{(x + 4)^2}}} =$

$\sqrt{1 + (x + 1)\sqrt{1 + (x + 2)\sqrt{1 + x^2 + 8x + 15}}} =$

$\sqrt{1 + (x + 1)\sqrt{1 + (x + 2)\sqrt{1 + (x + 3)(x + 5)}}} =$

$\sqrt{1 + (x + 1)\sqrt{1 + (x + 2)\sqrt{1 + (x + 3)\sqrt{(x + 5)^2}}}} =$

To show that this pattern continues indefinitely we just need to show:

$\sqrt{ 1 + (x + n)\sqrt{(x + n + 2)^2}} =$ $\sqrt{1 + (x + n)\sqrt{1 + (x + n + 1)(x + n + 3)}}$

but it's fairly obvious that $(x + n + 2)^2 = (x + n)^2 + 4(x + n) + 4$

and

$1 + (x + n + 1)(x + n + 3) = 1 + (x + n)^2 + 4(x + n) + 3 = (x + n)^2 + 4(x + n) + 4$

$\therefore f(x) = x + 2 = \sqrt{x + \sqrt{(x + 1)\sqrt{1 + (x + 2)\sqrt{1 + (x + 3)\sqrt{1 + ...}}}}}$

Let $w(x) = \dfrac{1}{f(x)} = \dfrac{1}{x + 2} \implies$

$\implies \displaystyle\int_{1}^{a} w(x) \:\ dx = \ln(x + 2)|_{1}^{a} = \ln(\dfrac{a + 2}{3}) =$

$2\ln(\dfrac{a}{\sqrt{3}}) = \ln(\dfrac{a^2}{3}) \implies a^2 -a - 2 = 0 \implies (a - 2)(a + 1) = 0$

and $a > 1 \implies a = \boxed{2}$ .