Answer
$F'(x) = \frac{x^2}{1+x^3}$
Work Step by Step
The function $f(t) = \frac{t^2}{1+t^3}$ is continuous on the interval $(0,\infty)$
$F(x) = \int_{0}^{x}f(t)~dt$
According to the Fundamental Theorem of Calculus (Part 1):
$F'(x) = f(x)$
Therefore:
$F'(x) = \frac{x^2}{1+x^3}$