Answer
$$\frac{1}{4}{\left( {\ln x} \right)^2} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\ln \sqrt x }}{x}} dx \cr
& {\text{Write }}\sqrt x {\text{ as }}{x^{1/2}} \cr
& \int {\frac{{\ln {x^{1/2}}}}{x}} dx \cr
& {\text{Using logarithmic properties}} \cr
& = \frac{1}{2}\int {\frac{{\ln x}}{x}} dx \cr
& {\text{Let }}u = \ln x,{\text{ }}du = \frac{1}{x}dx,{\text{ }}dx = xdu \cr
& \frac{1}{2}\int {\frac{{\ln x}}{x}} dx = \frac{1}{2}\int {\frac{u}{x}} xdu \cr
& = \frac{1}{2}\int u du \cr
& {\text{Integrating}} \cr
& = \frac{1}{2}\left( {\frac{{{u^2}}}{2}} \right) + C \cr
& = \frac{1}{4}{u^2} + C \cr
& {\text{Write in terms of }}x \cr
& = \frac{1}{4}{\left( {\ln x} \right)^2} + C \cr} $$
