Chapter 7 - Section 7.1 - The Logarithm Defined as an Integral - Exercises - Page 401: 16

$-2 e^{-\sqrt r} +c$

Solve $\int e^{-\sqrt r} \dfrac{1}{\sqrt r}dr$ Let us $p=-\sqrt r$ and $dp=-\dfrac{1}{2 \sqrt r}dr$ This implies $\int e^{-\sqrt r} \dfrac{1}{\sqrt r}dr=-2\int e^p dp $ $=-2 e^{p} +c$ $=-2 e^{-\sqrt r} +c$ Hence, $\int e^{-\sqrt r} \dfrac{1}{\sqrt r}dr=-2 e^{-\sqrt r} +c$