Q:

Following definite integral

Here is the integral:
$$\int_{0}^{2}\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}} dx$$
Here is my work: 
$$\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}} := y \implies x=y^2-y$$
By implicit differentiation, $$1 = 2y\frac{dy}{dx}-\frac{dy}{dx} \implies dx=dy(2y-1)$$.
So the integral is $$\int_{0}^{2}y(2y-1)dy = \frac{10}{3}$$.
(The limits of the integral stay at $0$ and $2$)
However, Wolfram Alpha is giving me approximately $19/6$; http://www.wolframalpha.com/input/?i=int%28%28x%2B%28x%2B%28x%2B%28x%2B%28x%2B%28x%2B%28x%2B%28x%2B%28x%2B%28x%2B%28x%2B%28x%2B%28x%2B%28x%2B%28x%29%5E%280.5%29%29%5E%280.5%29%29%5E%280.5%29%29%5E%280.5%29%29%5E%280.5%29%29%5E%280.5%29%29%5E%280.5%29%29%5E%280.5%29%29%5E%280.5%29%29%5E%280.5%29%29%5E%280.5%29%29%5E%280.5%29%29%5E%280.5%29%29%5E%280.5%29%29%5E%280.5%29%2C0%2C2%29. 
Is there something wrong with my work?

A:

Is there something wrong with my work?

Yes, you didn't change the integral limits correctly.
You have (see below)
$$\lim_{x\downarrow 0} \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\dotsc}}}} = 1,$$
so the integral should be
$$\int_1^2 y(2y-1)\,dy.$$
With $f(x) = \sqrt{x+\sqrt{x+\sqrt{x+\dotsc}}} = \sqrt{x+f(x)}$, for $x > 0$ we have $f(x) \geqslant 0$, whence $f(x) = \sqrt{x+f(x)} \geqslant \sqrt{x}$. Then $f(x) = \sqrt{x+f(x)} \geqslant \sqrt{f(x)} \geqslant \sqrt[4]{x}$, and iterating $f(x) \geqslant x^{1/2^k}$ for all $k \in\mathbb{N}$, which implies $f(x) \geqslant 1$.
From $x = y^2-y$ we can directly compute $y = f(x)$ with the quadratic formula and obtain
$$f(x) = \frac{1}{2} + \sqrt{x+ \frac{1}{4}},$$
which we can integrate to check the result.

