Answer
$$\frac{{2{{\left( {x + 1} \right)}^{3/2}}}}{3} - 2{\left( {x + 1} \right)^{1/2}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{x}{{\sqrt {x + 1} }}} dx \cr
& {\text{substitute }}u = x + 1,{\text{ }}x = u - 1,{\text{ }}du = dx, \cr
& = \int {\frac{x}{{\sqrt {x + 1} }}} dx = \int {\frac{{u - 1}}{{\sqrt u }}} du \cr
& \operatorname{simplify} \cr
& = \int {\frac{{u - 1}}{{{u^{1/2}}}}} du = \int {\left( {\frac{u}{{{u^{1/2}}}} - \frac{1}{{{u^{1/2}}}}} \right)} du \cr
& = \int {\left( {{u^{1/2}} - {u^{ - 1/2}}} \right)} du \cr
& {\text{find the antiderivative}} \cr
& = \frac{{{u^{3/2}}}}{{3/2}} - \frac{{{u^{1/2}}}}{{1/2}} + C \cr
& = \frac{{2{u^{3/2}}}}{3} - 2{u^{1/2}} + C \cr
& {\text{replace back}}\,\,\,u = x + 1 \cr
& = \frac{{2{{\left( {x + 1} \right)}^{3/2}}}}{3} - 2{\left( {x + 1} \right)^{1/2}} + C \cr} $$
