Answer
$$\frac{2}{3}{x^{3/2}}\ln x - \frac{4}{9}{x^{3/2}} + C$$
Work Step by Step
$$\eqalign{
& \int {\sqrt x \ln x} dx \cr
& {\text{radical property }}\sqrt x = {x^{1/2}} \cr
& \int {{x^{1/2}}\ln x} dx \cr
& {\text{substitute }}u = \ln x,{\text{ }}du = \frac{1}{x}dx \cr
& dv = {x^{1/2}}dx,{\text{ }}v = \frac{{{x^{3/2}}}}{{3/2}} = \frac{2}{3}{x^{3/2}} \cr
& {\text{ integration by parts}} \cr
& \int {udv} = uv - \int {vdu} \cr
& {\text{, we have}} \cr
& \int {{x^{1/2}}\ln x} dx = \ln x\left( {\frac{2}{3}{x^{3/2}}} \right) - \int {\left( {\frac{2}{3}{x^{3/2}}} \right)\left( {\frac{1}{x}} \right)dx} \cr
& \int {{x^{1/2}}\ln x} dx = \frac{2}{3}{x^{3/2}}\ln x - \frac{2}{3}\int {{x^{1/2}}dx} \cr
& {\text{find antiderivative}} \cr
& \int {{x^{1/2}}\ln x} dx = \frac{2}{3}{x^{3/2}}\ln x - \frac{2}{3}\left( {\frac{2}{3}{x^{3/2}}} \right) + C \cr
& \int {{x^{1/2}}\ln x} dx = \frac{2}{3}{x^{3/2}}\ln x - \frac{4}{9}{x^{3/2}} + C \cr} $$
