Answer
$2\sinh\sqrt{x}+C$
Work Step by Step
Use Th.5.18, Derivatives and Integrals of Hyperbolic Functions
$\displaystyle \int\cosh udu=\sinh u+C$
----
$\displaystyle \int\frac{\cosh\sqrt{x}}{\sqrt{x}}dx=\left[\begin{array}{lll}
u=\sqrt{x} & & \\
du=\frac{1}{2\sqrt{x}}dx. & \Rightarrow & 2\sqrt{x}du=dx
\end{array}\right]$
$=2\displaystyle \int\cosh\underbrace{(\sqrt{x})}_{u}\underbrace{(\frac{1}{2\sqrt{x}})dx}_{du}$
$=2\sinh\sqrt{x}+C$
