The value of the integral

$$ \int_1^{100} \frac{[x]}{x} d x $$

where $[x]$ is the greatest integer less than or equal to $x$ for any real number $x$, is

If $p(t)=\frac{t(t-1) \cdots(t-2019)}{2019!}$, then the value of

$$ \int_0^1\left(\frac{1}{t+1}+\frac{1}{t+2}+\cdots+\frac{1}{t+2020}\right) p(-t-1) d t $$

is:

$F(x)=\int_0^{e^x}\left(t^3+2 t^2-t-2\right) d t$, then for how many real numbers $x$ does $F^{\prime}(x)=0$ ?