This problem was just for fun
Here was what I managed to come up with:
The integral is approximately equal to $0.242070053984$.
The integral equals
$$\sum_{n=1}^{\infty}\int_{\sqrt{n}}^{\sqrt{n+1}}\frac{x-\sqrt{n}}{x}dx$$
$$\sum_{n=1}^{\infty}\int_{\sqrt{n}}^{\sqrt{n+1}}\left(1-\frac{\sqrt{n}}{x}\right)dx$$
$$\sum_{n=1}^{\infty}\left(\sqrt{n+1}-\sqrt{n}-\sqrt{n}\int_{\sqrt{n}}^{\sqrt{n+1}}\frac{1}{x}dx\right)$$
$$\sum_{n=1}^{\infty}\left(\sqrt{n+1}-\sqrt{n}-\sqrt{n}\ln\left(\sqrt{n+1}\right)+\sqrt{n}\ln\left(\sqrt{n}\right)\right)$$
I found many ways to manipulate this summation but I could not figure out whether any of them actually would make it easier to solve. For example, $\ln\left(\prod_{n=1}^{\infty}\left(\left(e^{\sqrt{\frac{n+1}{n}}-1}\sqrt{\frac{n}{n+1}}\right)^{\sqrt{n}}\right)\right)$ is not any easier to solve (I think).
And before you ask: No, I do not know whether or not this is actually solvable.
Thanks in advance.
$$a_n=\sqrt{n+1}-\sqrt{n}-\sqrt{n}\log\left(\sqrt{n+1}\right)+\sqrt{n}\log\left(\sqrt{n}\right)$$ Using Taylor series $$a_n=\frac 18 \sum_{k=1}^\infty (-1)^{k+1}\,\frac {\alpha_n}{\beta_n}\, n^{- \frac {2k+1} 2}$$ where the $\alpha_n$ form sequence $A120778$ in $OEIS$ and the first $\beta_n$ form the (unknown) sequence $$\{1,6,16,160,384,1792,4096,73728,163840,720896,1572864,13631488,\cdots\}$$ $$\sum_{n=1}^\infty a_n=\frac 18 \sum_{k=1}^\infty (-1)^{k+1}\,\frac {\alpha_n}{\beta_n}\,\, \zeta \left(\frac{2k+1}{2} \right)$$
Computing the partial sums (up to $p$) shows a very slow convergence $$\left( \begin{array}{cc} p & \sum_{n=1}^p a_n\\ 100 & 0.239326 \\ 200 & 0.240460 \\ 300 & 0.240849 \\ 400 & 0.241046 \\ 500 & 0.241165 \\ 600 & 0.241245 \\ 700 & 0.241302 \\ 800 & 0.241345 \\ 900 & 0.241379 \\ 1000 & 0.241406 \\ \end{array} \right)$$
Using Euler-MacLaurin summation formula $$\sum_{n=1}^\infty a_n=C-\frac{1}{4 \sqrt{n}}\left(1-\frac{19}{36 n}+\frac{49}{120 n^2}-\frac{6889}{20160 n^3}+\frac{5051}{17280 n^4}+O\left(\frac{1}{n^5}\right) \right)$$ for a suitable constant $C$.
For this level of expansion $$C=\frac{220266777361440211}{259003680974438400}-\frac{1262147487374869}{738871813865472 \sqrt{2}}+\frac{\pi }{6}+\frac{9293445205 \log (2)}{90194313216}$$