I need to find the general nth degree polynomial expansion of $x(x+1)(x+2)\dots(x+n)$ in order to solve an integral. Expanding $x(x+1)(x+2)\dots(x+n)$ has been challenging for me.
Say $n=5$, you will notice many patterns like $x^n+(n-1)x^{n-1}+\dotsb$ et cetera, but does there exist a general polynomial expansion for any $n$? This has been driving me nuts. I need a whiteboard.
The integral is $$\frac{1}{x(x+1)(x+2)\dots(x+n)}$$ and will be easy once I expand the polynomial. I am trying to avoid partial fraction decomposition.

$f(x) = \frac{1}{(x-a)(x-b)\cdots(x-k)} = \frac {A}{x-a} + \frac {B}{x-b} \cdots \frac {K}{(x-k)}$
$A = \lim_\limits {x\to a} (x-a)f(x) = \frac {1}{(a-b)(a-c)\cdots(a-k)}$
$f(x) = {1}{(x)(x+1)\cdots(x+n)} = \frac 1{n!}\frac {1}{x} - \frac {1}{(n-1)!}\frac {1}{x+1} + \frac {1}{2(n-2)!}\frac {1}{x+2} \cdots\\
f(x) = \sum_{i=0}^n\frac {(-1)^n}{i!(n-i)!} \frac{1}{x+i}$
By the way, supposing you had a nice way to expand the polynomial in the denominator, then what?  It still does not integrate without the partial fraction decomposition.
