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.

Let us find the partial fraction decomposition of the rational function
$$\frac{1}{x(x + 1)(x + 2) \cdots (x + n)}, \quad n = 0,1,2,\ldots$$
using the Heaviside cover-up method.
As we have $n + 1$ distinct linear factors we can write the rational function as
$$\frac{1}{x(x + 1)(x + 2) \cdots (x + n)} = \frac{A_0}{x} + \frac{A_1}{x + 1} + \frac{A_2}{x + 2} + \cdots + \frac{A_n}{x + n}.$$
To find each of the unknown constants $A_0$ through to $A_n$ we "cover-up" the first linear factor in the denominator on the left hand side then substitute in the value of $x$ that makes the factor covered up equal to zero which then gives the value for $A_0$. In the first case $x = 0$. We then repeat this process for each linear factor in the denominator. Doing so we find
\begin{align*}
A_0 &= \frac{1}{1.2.\ldots n} = \frac{1}{0! n!}\\
A_1 &= \frac{1}{(-1).1.2\ldots (n - 1)} = -\frac{1}{1!(n - 1)!}\\
A_2 &= \frac{1}{(-2)(-1).1.2\ldots (n - 2)} = \frac{1}{2!(n - 2)!}\\
& \vdots\\
A_n &= \frac{1}{(-1)(-n + 1)(-n + 2) \cdots (-1)} = \frac{(-1)^n}{n! 0!}
\end{align*}
So we see the $k$th constant $A_k$ where $0 \leqslant k \leqslant n$ will be given by
$$A_k = \frac{(-1)^k}{(n - k)! k!}, \quad 0 \leqslant k \leqslant n.$$
The partial fraction decomposition for the rational function will therefore be
$$\frac{1}{x(x + 1)(x + 2) \cdots (x + n)} = \sum_{k = 0}^n \frac{(-1)^k}{(n - k)! k!} \frac{1}{x + k}.$$
Now the integral you are after can be readily found. Here we have
$$\int \frac{dx}{x(x + 1)(x + 2) \cdots (x + n)} = \sum_{k = 0}^n \frac{(-1)^k}{(n - k)! k!} \ln |x + k| + C.$$
