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

Step $1$

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