Antiderivative of $x\sqrt{x+1}$

Let \(u=x+1\), so \(x=u-1\) and \(du=dx\). Then \[ (u-1)u^{1/2}=u^{3/2}-u^{1/2}, \] so \[ \int x\sqrt{x+1}\,dx =\frac25u^{5/2}-\frac23u^{3/2}+C. \] Therefore \[ \boxed{\int x\sqrt{x+1}\,dx=\frac25(x+1)^{5/2}-\frac23(x+1)^{3/2}+C.} \]