Evaluate the integral $\int _0^{\infty }\frac{dx}{(x+\sqrt{x\ +\ \sqrt{x\ +\ \sqrt{...}}})^2}$ 
Evaluate the integral $$\int _0^{\infty }\frac{dx}{\Big(x+\sqrt{x\ +\ \sqrt{x\ +\ \sqrt{...}}}\Big)^2}$$

How can we evaluate this integral? I've no idea what to do with this. At first, I thought of simplifying the denominator but that seems not working here.
Can anyone give me some hints?

Edit:
I've simplified the denominator and got:
$$\int \dfrac{4}{(2x + \sqrt{4x+1} + 1)^2}dx$$
What to do now?
 A: Let, \begin{align} \ x+ \sqrt {x+
{\sqrt {x + {\sqrt {...}}}}} = z^2
\Rightarrow {\frac{d}{dx}}(\ x +\sqrt {z^2})={\frac{d}{dx}}(\ z^2) \Rightarrow \ (1+{\frac{dz}{dx}})=\ 2z{\frac{dz}{dx}}
\Rightarrow {dz}= {\frac {1}{2z-1}}{\ dx}  \end{align}
Now by substituting,
\begin{align}  \int_{0}^{\infty} {\frac {dx}{\Big(x+\sqrt{x\ +\ \sqrt{x\ +\ \sqrt{...}}}\Big)^2}} = \int_{1}^{\infty} {\frac {2z-1}{z^4}}{\ dz} = {\frac {2}{3}}\end{align}
A: Begin by solving the quadratic equation for $y$:
$$y=x+\sqrt{y} \implies y=\frac{1}{2} \left(1+2x\pm \sqrt{1+4x}\right)$$
We will take the positive square root solution as to allow the antiderivative to converge at $0$.
$$\implies I=\int_{0}^{\infty} \frac{4}{(1+2x+\sqrt{1+4x})^2} \, dx$$
Now enforce the substitution $u^2=1+4x \> \implies 2u \, du = 4 \, dx$.
$$\implies I = \int_{1}^{\infty} \frac{8u}{\left(1 + 2u+ u^2\right)^2} \, du = \int_{1}^{\infty} \frac{8u}{\left(1+u\right)^4} \, du$$
Now write the $8u$ as $8u+8-8$ and split the integrand.
$$\implies I = \int_{1}^{\infty} \frac{8}{(u+1)^3} -\frac{8}{(u+1)^4} \, du = \frac{8}{3(u+1)^3} -\frac{4}{(u+1)^2} \bigg]_{1}^{\infty} =\frac{2}{3}$$
A: Under the change of variable $u=\sqrt{4x+1}$, then $x=\frac14(u^2-1),dx=\frac12udu$. So
\begin{eqnarray}
&&\int_0^\infty \dfrac{4}{(2x + \sqrt{4x+1} + 1)^2}dx\\
&=&\int_1^\infty \dfrac{4}{(\frac12(u^2-1) + u + 1)^2}\frac12udu\\
&=&8\int_1^\infty\frac{u}{(u^2+2u+1)^2}du\\
&=&8\int_1^\infty\frac{u}{(u+1)^4}du\\
&=&\frac32.
\end{eqnarray}
