It sounds like you're talking about an integral, namely
[tex]\displaystyle\int x^2\sqrt{x^3+9}\,\mathrm dx[/tex]
Substitute u = x ³ + 9, so that du = 3x ² dx. Then
[tex]\displaystyle\int x^2\sqrt{x^3+9}\,\mathrm dx=\frac13\int\sqrt{u}\,\mathrm du=\frac13\times\frac23u^{\frac32}+C=\boxed{\frac29(x^3+9)^{\frac32}+C}[/tex]
