Answer
$$\frac{2}{3}x{\left( {x - 1} \right)^{3/2}} - \frac{4}{{15}}{\left( {x - 1} \right)^{5/2}} + C$$
Work Step by Step
$$\eqalign{
& \int {x\sqrt {x - 1} } dx \cr
& {\text{Integrating by parts}} \cr
& u = x,{\text{ }}du = dx \cr
& dv = \sqrt {x - 1} dx,{\text{ }}v = \int {{{\left( {x - 1} \right)}^{1/2}}dx} ,{\text{ }}v = \frac{2}{3}{\left( {x - 1} \right)^{3/2}} \cr
& \int {udv} = uv - \int {vdu} \cr
& \int {x\sqrt {x - 1} } dx = \frac{2}{3}x{\left( {x - 1} \right)^{3/2}} - \int {\frac{2}{3}{{\left( {x - 1} \right)}^{3/2}}dx} \cr
& {\text{Integrating}} \cr
& = \frac{2}{3}x{\left( {x - 1} \right)^{3/2}} - \frac{2}{3}\int {{{\left( {x - 1} \right)}^{3/2}}dx} \cr
& = \frac{2}{3}x{\left( {x - 1} \right)^{3/2}} - \frac{2}{3}\left( {\frac{{{{\left( {x - 1} \right)}^{5/2}}}}{{5/2}}} \right) + C \cr
& = \frac{2}{3}x{\left( {x - 1} \right)^{3/2}} - \frac{4}{{15}}{\left( {x - 1} \right)^{5/2}} + C \cr} $$