## Calculus (3rd Edition)

$2e^{\sqrt x}+c$
Recall that $(e^x)'=e^x$ Let $u= \sqrt x$, then $du=\frac{1}{2\sqrt x} dx$ and hence we have $$\int \frac{e^{\sqrt x}}{\sqrt x} dx=2\int e^{u} du\\ =2e^u+c=2e^{\sqrt x}+c .$$