If you mean to find the antiderivative
[tex]\displaystyle\int x^2\sqrt{x+1}\,\mathrm dx[/tex]
substitute [tex]y=x+1[/tex], so that [tex]x=y-1[/tex] and [tex]\mathrm dx=\mathrm dy[/tex]. Then expand the squared term and distribute [tex]\sqrt y[/tex]:
[tex]\displaystyle\int(y-1)^2\sqrt y\,\mathrm dy=\int\left(y^{5/2}-2y^{3/2}+y^{1/2}\right)\,\mathrm dy[/tex]
[tex]=\dfrac27y^{7/2}-\dfrac45y^{5/2}+\dfrac23y^{3/2}+C[/tex]
[tex]=\dfrac2{105}y^{3/2}\left(15y^2-42y+35\right)+C[/tex]
[tex]=\dfrac2{105}(x+1)^{3/2}\left(15(x+1)^2-42(x+1)+35\right)+C[/tex]
[tex]=\dfrac2{105}(x+1)^{3/2}\left(15x^2-12x+8\right)+C[/tex]
Or if instead you mean to write
[tex]\displaystyle\int\frac{x^2}{\sqrt{x+1}}\,\mathrm dx[/tex]
the same substitution as before leads to
[tex]\displaystyle\int\frac{(y-1)^2}{\sqrt y}\,\mathrm dy=\int\left(y^{3/2}-2y^{1/2}+y^{-1/2}\right)\,\mathrm dy[/tex]
[tex]=\dfrac25y^{5/2}-\dfrac43y^{3/2}+2y^{1/2}+C[/tex]
[tex]=\dfrac2{15}y^{1/2}\left(3y^2-10y+15\right)+C[/tex]
[tex]=\dfrac2{15}(x+1)^{1/2}\left(3(x+1)^2-10(x+1)+15\right)+C[/tex]
[tex]=\dfrac2{15}(x+1)^{1/2}\left(3x^2-4x+8\right)+C[/tex]
