Let $f:(0,+\infty )\to \mathbb{R}$ and $F(x)={\int }_0^{x}f(t)dt$. If $F(x^2)=x^2(1+x)$, then $f(4)$ equals:

Given $F(x^2)={\int }_0^{x^2}f(t)dt=x^2+x^3$.Differentiating both sides with respect to $x$ using the Leibnitz Rule:
$$\frac{d}{dx}\left[{\int }_0^{x^2}f(t)dt\right]=\frac{d}{dx}(x^2+x^3)$$
$$f(x^2)\cdot \frac{d}{dx}(x^2)-f(0)\cdot \frac{d}{dx}(0)=2x+3x^2$$
$$f(x^2)\cdot 2x=2x+3x^2$$
Dividing by $2x$ (for $x\ne 0$):
$$f(x^2)=\frac{2x+3x^2}{2x}=1+\frac{3}{2}x$$
To find $f(4)$, set $x^2=4\Rightarrow x=2$ (since domain is positive):
$$f(4)=1+\frac{3}{2}(2)=1+3=4$$