How should I integrate this?

If you are familiarized with Taylor series, is easy to see that $$ e^{u(x)} = \sum_{k=0}^\infty \frac{u(x)^k}{k!}. $$ In your case, you must plug in your $u(x)$. Try to compute some terms and see that you will end up with the same summation as your integrand (avoiding the square root).

If you want to compute the integral using the complete expansion and not a truncated one (which it seems to be quite difficult or rather impossible to compute analitically), you will end up with $$ \int \frac{\mathrm{d}x}{\sqrt{e^{u(x)}}} \qquad \text{ with } \qquad u(x) = \left(\frac{x-35.8}{0.554}\right)^2, $$ so $e^{u(x)}$ looks like a gaussian function. Its indefinite integral can not be expressed in terms of elementary functions, but "we can deal with it" using the error function. The indefinite integral returns $$ \boxed{\int \frac{\mathrm{d}x}{\sqrt{e^{u(x)}}} = 0.554\sqrt{\frac{\pi}{2}}\mathrm{erf}(1.276 (x-35.8)) + C.}\tag{1}\label{eq:indefinite} $$

If you want to compute it with bounds setting a lower bound $x_0$, then $$ \boxed{\int_{x_0}^\infty \frac{\mathrm{d}x}{\sqrt{e^{u(x)}}} = 0.554\sqrt{\frac{\pi}{2}}\left[1-\mathrm{erf}(1.276(x_0-35.8))\right].} $$

Just for completness, this is how it looks \eqref{eq:indefinite} setting $C=0$.