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:

• A

  2019 .
• Correct answer
  B

  $2020^2$.
• C

  2020 .
• D

  $2019^2$.

Hence, the correct answer is option: 'B'.