Showing $\int_1^\infty\left(\sqrt{\sqrt{x}-\sqrt{x-1}}-\sqrt{\sqrt{x+1}-\sqrt{x}}\right)dx=\frac4{15}\left(\sqrt{26\sqrt2-14}-2\right)$ A Putnam problem asked to show that some improper  integral is convergent, but I was curious to see if it can be computed in closed form and Mathematica came up with this:
$$\int_1^{\infty} \left(\sqrt{\sqrt{x}-\sqrt{x-1}} -\sqrt{\sqrt{x+1}-\sqrt{x}      }     \right)dx=
\frac{4}{15} \left(\sqrt{26\sqrt{2}-14 }-2\right) \approx 0.739132$$
I did a few substitutions but it didn't  turn out as an easy calculation.  I remember that eliminating square roots required some Euler type substitutions. Any ideas of how one can arrive at such a surprising result?

This is a partial answer.
\begin{align}
a_n&=\int_n^{n+1} \left(\sqrt{\sqrt{x}-\sqrt{x-1}} -\sqrt{\sqrt{x+1}-\sqrt{x}}     \right)\mathbb dx\\
&=\int_n^{n+1} \sqrt{\sqrt{x}-\sqrt{x-1}}\,\mathbb dx-\int_{n+1}^{n+2} \sqrt{\sqrt{x}-\sqrt{x-1}}\mathbb\,dx\\
\sum_{n=1}^{\infty}a_n&=\int_1^{2} \sqrt{\sqrt{x}-\sqrt{x-1}}\,\mathbb dx
=\int_1^{2} \frac1{\sqrt{\sqrt{x}+\sqrt{x-1}}}\,\mathbb dx
\end{align}
The WolframAlpha shows that the indefinite integral is
$$\int \frac1{\sqrt{\sqrt{x}+\sqrt{x-1}}}\,\mathbb dx=\frac{5\left(\sqrt{x-1}+\sqrt x\right)^4+3}{15\left(\sqrt{x-1}+\sqrt x\right)^{5/2}}+C$$
, which I didn't figure out how to get there.

I completed the answer below.
\begin{align}
\sqrt{x}+\sqrt{x-1}&\to t\\
dt&=\frac12\left(\frac1{\sqrt{x}}+\frac1{\sqrt{x-1}}\right)dx=\frac12\frac{t}{\sqrt{x}\sqrt{x-1}}dx\\
t^2&=2x-1+2\sqrt x\sqrt{x-1}\\
4x(x-1)&=(t^2-2x+1)^2=t^4-2(2x-1)t^2+(2x-1)^2\\
0&=t^4-4xt^2+2t^2+1\\
x&=\frac{(t^2+1)^2}{4t^2}\\
x-1&=\frac{(t^2-1)^2}{4t^2}\\
\int \frac1{\sqrt{t}}\,\mathbb dx&=\int \frac{2\sqrt{x}\sqrt{x-1}}{t^{3/2}}\,\mathbb dt=\int \dfrac{2\dfrac{t^4-1}{4t^2}}{t^{3/2}}\mathbb dt\\
&=\frac12 \int \left(t^{1/2}-t^{-7/2}\right)\mathbb dt\\
&=\frac13t^{3/2}+\frac15t^{-5/2}+C\\
&=\frac13\left(\sqrt{x}+\sqrt{x-1}\right)^{3/2}+\frac15\left(\sqrt{x}+\sqrt{x-1}\right)^{-5/2}+C
\end{align}
