Evaluate the integral $by$ making the given substitution.

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{x}^2\sqrt[\phantom{2}]{x^3+8}dx,u=x^3+8$

Step $1$

$We$ know that $ifu=f(x),$ then $du=f^{\text{'}}(x)dx.$ Therefore, $ifu=x^3+8,$ then $du=$

$3x^2$

$dx.$

Step $2$

$Ifu=x^3+8is$ substituted into ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{x}^2\sqrt[\phantom{2}]{x^3+8}dx,$ then $we$ have ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{x}^2(u{)}^{\frac{1}{2}}dx={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{u}^{\frac{1}{2}}{x}^{2}dx.$

$We$ must also convert $x^{2}dx$ into $an$ expression involving $u.$

$We$ know that $du=3x^{2}dx,$ and $so{x}^{2}dx=$

$du.$