This Question is an Extension of this Previously Asked Question: Nested root integral $\int_0^1 \frac{dx}{\sqrt{x+\sqrt{x+\sqrt{x}}}}$
I was looking into answering the question of whether it was possible to integrate the fully nested root integral of the variety described in the previous problem:
$$
\int \frac{dx}{\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}}
$$
So I started by defining the nested root in another way
$$
u=\sqrt{x+u} \therefore
\\
u^2-u=x
\\
(2u-1)du = dx
$$
Using the results of the substitution I have set up
$$
\int \frac{dx}{\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}} = \int \frac{2u-1}{u} du
\\
= \int \left(2-\frac{1}{u}\right)du = 2u-\ln(u)
$$
I am unsure of my work, as I have never attempted to integrate any infinitely nested functions. Therefore I have no idea whether my method for u-substitution is valid. Am I just living under a rock or have other people seen this method used previously? For such a seemingly intimidating problem it was surely quite easy.
I had difficulty checking my work using wolfram alpha, but I managed to confirm that this works for the definite integral limits from $x = 1$ to $x = 2$ and from $x = 1$ to $x = 3$. Maybe I am just flat out wrong and got lucky on these two calculations?
Short Answer:
Your work is perfectly fine if your lower and upper integral limits satisfy $0 < a \leq b$. In that case your answer $2u - \ln(u)$ even has a nice closed form purely in terms of $x$:
$$
\int_a^b \frac{dx}{\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}} = \Big[\big(1 + \sqrt{4x + 1}\big) - \ln\Big(\frac{1}{2}\big(1 + \sqrt{4x + 1}\big)\Big)\Big]_a^b
$$
This formula also continues to work for a lower limit of $a = 0$ if you interpret either the integral and/or the nested radical $\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}$ in the denominator properly enough.
Long Answer (Analysis):
Since you used $u$-substitution, your method should work as long as the conditions for an integration by substitution are met. Say you are integrating over some interval $[a, b]$. You have to verify:
Does the function $u(x) = \sqrt{x + u(x)}$ that you defined implicitly actually make sense over $[a, b]$? In other words, is there really a function $u : [a, b] \to \Bbb R$ that satisfies that recursion?
Is the function $u(x)$ actually differentiable over $[a, b]$?
$\underline{\textit{There is some good news for these questions:}}$
As long as $x > 0$, there is a well-defined expression for $u$ in terms of $x$ when $u = \sqrt{x + u}$. To realize this, we need translate the intuitive expression $\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}$ into the precise language of calculus. Only then can we bring the full power of calculus to bear on this problem. So formally what is going on with a nested radical like $\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}$ is this:
Let $u_{x,1} = \sqrt{x}$ and define recursively the sequence $u_{x,n + 1} = \sqrt{x + u_{x,n}}$ ($n \in \Bbb Z_+$). If $$u_x = \lim\limits_{n \to \infty}u_{x,n}$$ exists, then we may define our sought-after function $u$ at $x$ to be $u(x) = u_x$. In essence, the limit $\lim\limits_{n \to \infty}u_{x,n}$ is mathematically what we define the expression $\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}$ to be. And we can easily check that $u_x = \sqrt{x + u_x}$ by taking the limit as $n \to \infty$ at both sides of the equation $u_{x,n + 1} = \sqrt{x + u_{x,n}}$.
Now the good news is that, as long as $x > 0$, you can show that the sequence $u_{x,n}$ is bounded and monotonically increasing so that it does converge to a definite limit, namely
$$u_x = \frac{1}{2}\big(1 + \sqrt{4x + 1}\big)$$
Hence, our function $u(x) = u_x$ is well-defined for $x > 0$. Also, note that the formula above should not surprise you. You can easily see where it originated:
Informally, if you take your substitution equation $u^2 - u = x$ and wrote it as a quadratic equation $u^2 - u - x = 0$, you can solve it by thinking of $x$ as a constant. And indeed, one of the solutions that pops out is precisely $u_+ = \frac{1}{2}\big(1 + \sqrt{4x + 1}\big)$. You can eliminate the other solution $u_- = \frac{1}{2}\big(1 - \sqrt{4x + 1}\big)$ since it is negative if $x > 0$ and by convention square roots are positive.
So as long as $x > 0$, you can safely take $$u(x) = \frac{1}{2}\big(1 + \sqrt{4x + 1}\big)$$ as the $u$-substitution function which satisfies $u = \sqrt{x + u}$. In fact, as is apparent from the formula, $u(x)$ is even differentiable in this case.
$\underline{\textit{But there are caveats:}}$
$1.\ \textbf{Note that for $x < 0$, the limit does not make sense:}$ as the very first sequence element $u_{x,1} = \sqrt{x}$ is not real. So, from this very analysis, you can immediately conclude that you should not be integrating over negative values in your integral.
$2.\ \textbf{Next, at $x = 0$, things almost work out but break down anyway:}$ Note, we only managed to eliminate $\frac{1}{2}\big(1 - \sqrt{4x + 1}\big)$ as a candidate for the limit above because it was negative if $x > 0$. Well, if $x = 0$, then $u_- = \frac{1}{2}\big(1 - \sqrt{4x + 1}\big) = 0$ and you can no longer eliminate it that easily. So, we must go back to our definition of $\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}$ in terms of sequences to arbitrate between $u_+$ and $u_-$. Applying that definition when $x = 0$, we see that $u_- = 0$ is the candidate that is chosen this time not $u_+ = 1$. This is because in this case, all the sequence elements $u_{x,n}$ are zero: $$u_{x=0,1} = \sqrt{x} = \sqrt{0} = 0,\quad u_{x=0,2} = \sqrt{x + u_{x=0,1}} = \sqrt{0 + 0} = 0,\quad \ldots \text{ etc}$$
Hence, $\lim\limits_{n \to \infty}u_{x=0,n} = 0$ and $u(0) = 0$. However, approaching $0$ from the right, we see that
$$\lim\limits_{x \to 0+}u(x) = \frac{1}{2}\big(1 + \sqrt{4\cdot0 + 1}\big) = 1$$ And therefore even though $u(x)$ is defined at $x = 0$, it is sadly not continuous there, let alone differentiable. So the $u$-substitution Theorem no longer applies.
In any case, there is an even worse problem when $x = 0$. Note that the function you are trying to integrate $f(x) = \sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}$ is undefined at $x = 0$ because as we saw, our definition of $\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}$ in terms of sequences gives you a $0$ when $x = 0$. So there would be a $0$ in your denominator for $f(x)$ if that was allowed.
$\underline{\textit{Okay, so we have concluded so far that:}}$
As long as your integration interval $[a, b]$ satisfies $0 < a \leq b$, your work should go through and you can use $u(x) = \frac{1}{2}\big(1 + \sqrt{4x + 1}\big)$ as the explicit formula for $u$ to express your final integral answer:
$$
\int_a^b \frac{dx}{\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}} = \Big[\big(1 + \sqrt{4x + 1}\big) - \ln\Big(\frac{1}{2}\big(1 + \sqrt{4x + 1}\big)\Big)\Big]_a^b
$$
$\underline{\textit{Fixing the breakdown at $x = 0$:}}$
If you really want $x = 0$ as one of the limits e.g.
$$
\int_0^b \frac{dx}{\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}}
$$ for $b > 0$, you can do so in two ways, both of which lead to the same result:
You can modify the definition of $\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}$ thus: it defaults to the usual definition via sequences if $x > 0$ and to $\frac{1}{2}(1 + \sqrt{4 \cdot 0 + 1}) = 1$ if $x = 0$. Then you can safely use $u(x) = \frac{1}{2}\big(1 + \sqrt{4x + 1}\big)$ for all $x \geq 0$. And the answer you will get for your integral is exactly what you would expect by plugging in $a = 0$ in the closed form I gave above:
$$
\big(1 + \sqrt{4b + 1}\big) - \ln\Big(\frac{1}{2}\big(1 + \sqrt{4b + 1}\big)\Big) - 2
$$
On the other hand, you can instead take a limiting integral in the same spirit that we define $\int_0^b \frac{1}{x^2}dx$ to get around the singularity of $\frac{1}{x^2}$ at $0$. That is, you can define:
\begin{align*}
\int_0^b \frac{dx}{\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}} &:= \lim_{a \to 0+}\int_a^b \frac{dx}{\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}} \\
&= \lim_{a \to 0+}\big[2u(x) - \ln(u(x))\big]_a^b
\end{align*} This leads to the same answer because ultimately $\lim\limits_{a \to 0^+}u(x) = 1$.
