Let $n$ integer $\geq 2$ ( or suffisemment big) fixed and $0<x<1$ fixed number

i'm looking for an integral representation of type $ \int_{0}^1 f^n(x).g(x)dx$ for $I_n$ where $ I_n=\displaystyle \sum_{k=0}^{n-1} C_{n+k}^n (-x)^k \alpha_{n,k}$ and where $\forall k$ such that $ 0 \leq k \leq n-1 $ $$ \alpha_{n,k}= \displaystyle \sum_{p=1}^{n-k} \displaystyle C_{n}^{n-k-p} \frac{(-1)^{p+1}}{p}$$.

i'm looking too for a sharp bound to the following expression $|I_n|$

I don't konw if it's usufull or not but it seems that NUMERICALLY $\forall k$ such that $ 0 \leq k \leq n-1 $ $ \alpha_{n,k} \geq 1 $ so positif and for n fixe $((n-k)! \alpha_{n,k})_{0\leq k \leq n-1} $ is an integer and a deceasing sequence in $ k$, i have no explanation for that.

I have found the following bound but it doest not interess me $$ \displaystyle |\sum_{k=0}^{n-1} C_{n+k}^n (-x)^k \alpha_{n,k}| \leq \displaystyle \sum_{k=0}^{n-1} C_{n+k}^n |x|^k |\alpha_{n,k}| \leq \displaystyle \sum_{k=0}^{n-1} C_{n+k}^n x^k 2^n \leq \displaystyle \sum_{k=0}^{+\infty} C_{n+k}^n (x)^k 2^n = \frac{2^n}{(1-x)^{n+1}} $$

if any one suggest an idea thanks for his help