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?

To compute \begin{align}I=\int_{0}^{1}f(x)\;dx\end{align} with $f(x)=\sqrt{\sqrt{x+1}-\sqrt{x}}$  you can use change of variables: $x=\cot^2(u)$ then $I=\int_{\pi/8}^{\pi/4} g(u) du$ where \begin{align}g(u)=2\cot(2u)\left(2+2 \cot^2 ( 2u)   \right) \sqrt {{\frac {1-\cos(2u)}{\sin \left( 2\,u \right) }}}=4{\frac {\cos \left( 2\,u \right) }{ \sin^3(2u)}\sqrt {{\frac {\sin \left( u \right) }{\cos
 \left( u \right) }}}}
\end{align}.
Now another change of variables, $u=\arctan(v^2)$ makes
\begin{align}
I=\int_{\sqrt{\sqrt2 -1}}^{1} \frac{(1-v^4)(1+v^4)}{v^4}  \;dv
=-{\frac{8}{15}}+{\frac { \left( 4\,\sqrt {2}+24 \right) \sqrt {\sqrt {
2}-1}}{15}}
\end{align}
since $\tan\frac\pi 8=\tan\left(\frac 12 \frac \pi4\right)=\sqrt2 -1$.
