The integration 3

With the substitution $x = u^2$ we obtain $\begin{aligned} \int_0^\infty \frac{\cosh(3\sqrt{x})}{\sqrt{x} e^x}\,dx & = \; 2\int_0^\infty e^{-u^2}\cosh(3u)\,du \; = \; \int_0^\infty e^{-u^2}\big(e^{3u} + e^{-3u}\big)\,du \; = \; \int_{-\infty}^\infty e^{-u^2 + 3u}\,du \\ & = \; \int_{-\infty}^\infty \mathrm{exp}\big[-(u-\tfrac32)^2 + \tfrac94\big]\,dx \; = \; e^{\frac{9}{4}} \int_{-\infty}^\infty e^{-u^2}\,du \; = \; \boxed{e^{\frac{9}{4}}\sqrt{\pi}} \end{aligned}$