Answer
$$
F(x)=\int_{0}^{x} \frac{t^{2}}{1+t^{3}} d t.
$$
The derivative of the function $F$ is given by:
$$
F^{\prime}(x)=\frac{d}{d x} \int_{0}^{x} \frac{t^{2}}{1+t^{3}} d t=\frac{x^{2}}{1+x^{3}}$$
Work Step by Step
$$
F(x)=\int_{0}^{x} \frac{t^{2}}{1+t^{3}} d t.
$$
The derivative of the function $F$ is given by:
$$
F^{\prime}(x)=\frac{d}{d x} \int_{0}^{x} \frac{t^{2}}{1+t^{3}} d t=\frac{x^{2}}{1+x^{3}}$$
