Answer
$$36.$$
Work Step by Step
By making use of (FTC II), we have $$ G'(x)=\frac{d}{d x} \int_{0}^{x^3} \sqrt{t+1} dt =\left(\frac{d}{d u} \int_{0}^{u}\sqrt{t+1} dt\right)\frac{du}{dx}=\sqrt{u+1} ( 3 x^2)\\ =3x^2 \sqrt{x^3+1}. $$ Hence, $$G'(x)=3(2^2) \sqrt{8+1}=36.$$
