$\begin{align}  & \int{x\sqrt{x+1}dx} \\ & \text{Let }{{u}^{2}}=x+1,\text{  }x={{u}^{2}}-1,\text{  }\frac{dx}{du}=2u,\text{ }dx=2udu \\ & \int{x\sqrt{x+1}}dx=\int{({{u}^{2}}-1)u(2udu)}=2\int{({{u}^{2}}-1){{u}^{2}}du} \\ & \int{x\sqrt{x+1}dx}=2\int{({{u}^{4}}-{{u}^{2}})du}=2\left[ \frac{{{u}^{5}}}{5}-\frac{{{u}^{3}}}{3} \right]+C \\ & \int{x\sqrt{x+1}dx}=2\left[ \frac{{{(x+1)}^{\tfrac{5}{2}}}}{5}-\frac{{{(x+1)}^{\tfrac{3}{2}}}}{3} \right]+C \\\end{align}$