How strongly does $\int_1^X \frac{\exp(B/\sqrt x)}{\sqrt x} dx$ depend on $B$? As part of analysing an algorithm I stumbled upon the integral $\displaystyle \int_1^X \frac{\exp(B/\sqrt x)}{\sqrt x} dx$ as an approximation for the corresponding sum. Wolfram alpha gives the antiderivative in terms of non-elementary functions. This seems more trouble than it's worth since I'm only interested in a readable bound for the integral.
One very poor bound  comes from rounding the exponential up to $e^B$ and get $\displaystyle e^B \int_1^X \frac{dx}{\sqrt x}  = 2e^B(\sqrt X - 1) = O(e^B \sqrt X) $. 
One better bound comes from breaking the intgral into two parts like this: 
$$\displaystyle \int_1^X \frac{\exp(B/\sqrt x)}{\sqrt x} dx= \int_1^{B^2}\frac{\exp(B/\sqrt x)}{\sqrt x} dx + \int_{B^2}^X \frac{\exp(B/\sqrt x)}{\sqrt x} dx$$
For the first part do the same rounding up to get
$$  \int_1^{B^2}\frac{\exp(B/\sqrt x)}{\sqrt x}\le e^B \int_1^{B^2}\frac{dx}{\sqrt x} \le 2e^B \sqrt {B^2} = 2B e^B.$$
For the second part the denominator is small and we get 
$$ \int_{B^2}^X \frac{\exp(B/\sqrt x)}{\sqrt x} dx \le \int_{B^2}^X \frac{e}{\sqrt x} dx = 2 e (\sqrt X - \sqrt {B^2}) \le 2e \sqrt X.$$
Putting it back together we can bound the original integral by $2Be^B + 2e \sqrt X$. We still have the exponential term but not it is a constant and not multiplied by the $\sqrt X$.
I wonder can we do any better than this? Can we replace the dependence on $B$ with something weaker?

You can easily find
$$
\int\frac{\exp(B/\sqrt{x})}{\sqrt{x}}dx=2\sqrt{x}\exp\left(\frac{B}{\sqrt{x}}\right)-2B\cdot\textrm{Ei}\left(\frac{B}{\sqrt{x}}\right),
$$
where
$$
\textrm{Ei}(z)=-\int^{\infty}_{-z}\frac{e^{-t}}{t}dt
$$
and then see at Wikipedia Ei
