We fix $x > 0$ and for all natural integers $n$, we define $F_{n}(x)$ by:
$$F_{n}(x) = \ln\left(\frac{n! \, n^{x+1}}{(x+1)(x+2) \ldots (x+n+1)}\right)$$
Deduce that:
$$F_{n}(x) = G_{n}(x) - \int_{0}^{n+1} \frac{h(u)}{u+x} du$$
where
$$G_{n}(x) = \ln n! + (x+1)\ln n - \left(x+n+\frac{3}{2}\right)\ln(x+n+1) + n+1 + \left(x+\frac{1}{2}\right)\ln x$$