<s>[INST] Integral of $n$-th Bernoulli polynomial of the second kind We have
\begin{align}
\int x(x−1)(x−2)...(x−n)\,dx=(n+1)!\cdot\psi_{n+2}(x),
\end{align}
where, $\psi_n(x)$ is the $n$-th Bernoulli polynomial of the second kind. We have
\begin{align}
    I &= \int_0^{n} \underbrace{x}_{\color{red}{=\,u}}\cdot \underbrace{x(x-1)\cdots(x-n)}_{\color{red}{=\,v'}}\,d u \\
        &= x\cdot (n+1)!\,\psi_{n+2}(x)\,\Big|_0^{n} - (n+1)! \int_0^{n} \psi_{n+2}(u)\,d u \\
        &= (n+1)!\left[n\cdot\psi_{n+2}(n) - \int_0^n \psi_{n+2}(u)\,d u\right] \\
        &= -(n+1)!\left[n\cdot|G_{n+2}| + \int_0^n \psi_{n+2}(u)\,d u\right]
\end{align}
What is $\displaystyle\int_0^n \psi_{n+2}(u)\,d u$ ?
 [/INST] It was so simple:
\begin{align}
     (n+2)!\cdot\psi_{n+3}(x) &= \int x(x-1)(x-2)\cdots(x-n)(x-(n+1))\ dx \\
     &=\underbrace{\int x \cdot x(x-1)(x-2)\cdots(x-n) \ dx}_{\color{red}{=\,I}} - (n+1) \underbrace{\int x(x-1)(x-2)\cdots(x-n) \ dx}_{\color{red}{=\,(n+1)!\,\cdot\,\psi_{n+2}(x)}} \\
&= I - (n+1)\cdot(n+1)!\,\cdot\,\psi_{n+2}(x)
\end{align}
Thus applies:
\begin{align}
     I &= (n+1)\cdot(n+1)!\,\cdot\,\psi_{n+2}(x) + (n+2)! \cdot \psi_{n+3}(x) \color{white}{\frac{1}{2}} \\
&= (n+1)! \cdot \left[(n+1)\cdot \psi_{n+2}(x) + (n+2)\cdot \psi_{n+3}(x)\right] \color{white}{\frac{1}{2}}
\end{align}
Now the only question is what is $\psi_{n+3}(n)$....
</s>