Evaluating $\int\frac{\mathrm dx}{\sqrt{\lfloor 1+ \sqrt{1+x}\rfloor}}$ How can I solve this integral?
$$\int\frac{\mathrm dx}{\sqrt{\lfloor 1+ \sqrt{1+x}\rfloor}}$$

Notice for any integer $n \ge 2$ and $x \ge 0$,
$$\begin{align}
\left\lfloor 1 + \sqrt{1+x}\right\rfloor = n
&\iff n \le 1 + \sqrt{1+x} < n+1\\
&\iff x \in [ (n-1)^2 - 1, n^2 - 1 )
\end{align}$$
To perform the integral over some interval $[0,X]$ where $X > 0$, you just need to split the interval $[0,X]$ to intervals of the form $[(n-1)^2-1, n^2 - 1)$. Let $N = \left\lfloor 1 + \sqrt{1+X}\right\rfloor$, we have
$$\begin{align}
\int_0^X \frac{dx}{\sqrt{\left\lfloor 1 + \sqrt{1+x}\right\rfloor}}
&= \left( \sum_{n=2}^{N-1}\int_{(n-1)^2-1}^{n^2-1} + \int_{(N-1)^2-1}^X\right)
\frac{dx}{\sqrt{\left\lfloor 1 + \sqrt{1+x}\right\rfloor}}\\
&= \sum_{n=2}^{N-1}\frac{2n-1}{\sqrt{n}} + \frac{X - (N-1)^2 + 1}{\sqrt{N}}
\end{align}
$$
