Answer
$$\frac{2}{3}{\left( {x + 1} \right)^{3/2}} - 2\sqrt {x + 1} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{x}{{\sqrt {x + 1} }}} dx \cr
& {\text{Let }}u = {\left( {x + 1} \right)^{1/2}}{\text{ }} \Rightarrow {\text{ }}{u^2} = x + 1{\text{ }} \Rightarrow {\text{ }}2udu = dx \cr
& {\text{Apply the substitution}} \cr
& \int {\frac{x}{{\sqrt {x + 1} }}} dx = \int {\frac{{{u^2} - 1}}{u}} \left( {2udu} \right) \cr
& {\text{ }} = 2\int {\left( {{u^2} - 1} \right)} du \cr
& {\text{Integrating}} \cr
& {\text{ }} = 2\left( {\frac{1}{3}{u^3} - u} \right) + C \cr
& {\text{ }} = \frac{2}{3}{u^3} - 2u + C \cr
& {\text{Substitute - back }}u = {\left( {x + 1} \right)^{1/2}} \cr
& {\text{ }} = \frac{2}{3}{\left( {{{\left( {x + 1} \right)}^{1/2}}} \right)^3} - 2{\left( {x + 1} \right)^{1/2}} + C \cr
& {\text{ }} = \frac{2}{3}{\left( {x + 1} \right)^{3/2}} - 2\sqrt {x + 1} + C \cr} $$
