Answer
$$2\sqrt r - 2\ln \left( {\sqrt r + 1} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dr}}{{1 + \sqrt r }}} \cr
& {\text{Integrate by using the substitution method}} \cr
& \,\,\,{\text{Let }}r = {u^2},\,\,\,\,dr = 2udu \cr
& \,\,\,\int {\frac{{dr}}{{1 + \sqrt r }}} = \int {\frac{{2u}}{{1 + \sqrt {{u^2}} }}} du \cr
& = \int {\frac{{2u}}{{1 + u}}} du \cr
& {\text{By the long division }}\frac{{2u}}{{1 + u}} = 2 - \frac{2}{{u + 1}} \cr
& = \int {\left( {2 - \frac{2}{{u + 1}}} \right)} du \cr
& {\text{Integrating}} \cr
& = 2u - 2\ln \left| {u + 1} \right| + C \cr
& \cr
& {\text{Write in terms of }}r,{\text{ }}r = {u^2},\,\,\,\sqrt r = u \cr
& = 2\sqrt r - 2\ln \left| {\sqrt r + 1} \right| + C \cr
& = 2\sqrt r - 2\ln \left( {\sqrt r + 1} \right) + C \cr} $$
