How to integrate $\int_{0}^{2} \sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}\,dx$? I want to integrate this $$\int_{0}^{2} \sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}\,dx$$ 

I tried making the substitution $u^2=x+u$ and  my integral turns into 

$$\int_{0}^{2}u(2u-1)\,du = \frac{10}{3}$$ 

However the answer is supposed to be $\dfrac{19}{6}$, I would like to know where I am making a mistake or another way
Let $L(x)=\sqrt{x+\sqrt{x+\cdots}}$.

$L(x)^2=x+L(x)$, hence, $L(x)^2-L(x)-x=0$ and $L(x)=\frac{1\pm\sqrt{1+4x}}{2}$.

Because $L(x)\ge0$, $L(x)=\frac{1+\sqrt{1+4x}}{2}$ and then the intergal is straightforward.
Hint:
solve the two equations to find the new limits of integration( use the $u^2-u=x$)
$$u^2-u=0$$
and
$$u^2-u=2$$