By parts:
[tex]u=x\implies\mathrm du=\mathrm dx[/tex]
[tex]\mathrm dv=\sqrt{x+1}\,\mathrm dx\implies v=\dfrac23(x+1)^{3/2}[/tex]
[tex]\displaystyle\int x\sqrt{x+1}\,\mathrm dx=\frac23x(x+1)^{3/2}-\frac23\int(x+1)^{3/2}\,\mathrm dx[/tex]
[tex]\displaystyle\int x\sqrt{x+1}\,\mathrm dx=\frac23x(x+1)^{3/2}-\frac4{15}(x+1)^{5/2}+C[/tex]
[tex]\displaystyle\int x\sqrt{x+1}\,\mathrm dx=\frac1{15}(x+1)^{3/2}\left(10x-4(x+1)\right)+C[/tex]
[tex]\displaystyle\int x\sqrt{x+1}\,\mathrm dx=\frac2{15}(x+1)^{3/2}\left(3x-2)\right)+C[/tex]
By substitution:
[tex]u=x+1\implies\mathrm du=\mathrm dx[/tex]
[tex]\displaystyle\int x\sqrt{x+1}\,\mathrm dx=\int(u-1)\sqrt u\,\mathrm du=\int u^{3/2}-u^{1/2}\,\mathrm du=\frac25u^{5/2}-\frac23u^{3/2}+C[/tex]
[tex]\displaystyle\int x\sqrt{x+1}\,\mathrm dx=\frac25(x+1)^{5/2}-\frac23(x+1)^{3/2}+C[/tex]
[tex]\displaystyle\int x\sqrt{x+1}\,\mathrm dx=\frac2{15}(x+1)^{3/2}\left(3(x+1)-5\right)+C[/tex]
[tex]\displaystyle\int x\sqrt{x+1}\,\mathrm dx=\frac2{15}(x+1)^{3/2}\left(3x-2\right)+C[/tex]
As you can see, no difference (as long as you simplify everything in a consistent way).
