Answer
$$2\sqrt x \ln x - 4\sqrt x + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\ln x}}{{\sqrt x }}dx} \cr
& {\text{radical property }}\sqrt x = {x^{1/2}} \cr
& \int {\frac{{\ln x}}{{{x^{1/2}}}}dx} \cr
& \int {{x^{ - 1/2}}\ln xdx} \cr
& {\text{substitute }}u = \ln x,{\text{ }}du = \frac{1}{x}dx \cr
& dv = {x^{ - 1/2}}dx,{\text{ }}v = \frac{{{x^{1/2}}}}{{1/2}} = 2{x^{1/2}} \cr
& {\text{ integration by parts}} \cr
& \int {udv} = uv - \int {vdu} \cr
& {\text{we have}} \cr
& \int {\frac{{\ln x}}{{\sqrt x }}dx} = 2{x^{1/2}}\left( {\ln x} \right) - \int {\left( {2{x^{1/2}}} \right)\left( {\frac{1}{x}} \right)dx} \cr
& \int {\frac{{\ln x}}{{\sqrt x }}dx} = 2\sqrt x \ln x - 2\int {{x^{ - 1/2}}} dx \cr
& {\text{find antiderivative}} \cr
& \int {\frac{{\ln x}}{{\sqrt x }}dx} = 2\sqrt x \ln x - 2\left( {2{x^{1/2}}} \right) + C \cr
& \int {\frac{{\ln x}}{{\sqrt x }}dx} = 2\sqrt x \ln x - 4\sqrt x + C \cr} $$
