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