Take the integral:
\[Integral]E^Sqrt[r]/Sqrt[r] \[DifferentialD]r
For the integrand E^Sqrt[r]/Sqrt[r], substitute u==Sqrt[r] and \[DifferentialD]u==1/(2 Sqrt[r])\[ThinSpace]\[DifferentialD]r:
== 2\[Integral]E^u \[DifferentialD]u
The integral of E^u is E^u:
== 2 E^u+constant
Substitute back for u==Sqrt[r]:
Answer: \[SpanFromLeft]
== 2 E^Sqrt[r]+constant - Courtesy of Mathematica 11 Home Edition
