Integrate $\int \prod_{i=0}^n \frac{1}{(x+i)}dx$ I initially tried to find a pattern using partial fraction decomposition:
$$
n= 1 \, : \, \, \frac{1}{x(x+1)} = \frac{1}{x}+\frac{-1}{x+1}
$$
$$
n=2 \, : \, \, \frac{1}{x(x+1)(x+2)} = \frac{1/2}{x}+\frac{-1}{x+1}+\frac{1/2}{x+2}
$$
$$
n=3 \, : \, \, \frac{1}{x(x+1)(x+2)(x+3)} = \frac{1/6}{x}+\frac{-1/2}{x+1}+\frac{1/2}{x+2} + \frac{-1/6}{x+3}
$$
$$
n=4 \, : \, \, \frac{1}{x(x+1)(x+2)(x+3)(x+4)} = \frac{1/24}{x}+\frac{-1/6}{x+1}+\frac{1/4}{x+2} + \frac{-1/6}{x+3} + \frac{1/24}{x+4}
$$
But I don't seem to find a pattern. Is there another method to resolve this? Thank you.

Hint. If we have that
$$\prod_{i=0}^n \frac{1}{(x+i)}=\sum_{i=0}^n \frac{A_i}{(x+i)}.$$
then, for $k=0,\dots,n$,
$$\lim_{x\to -k}(x+k)\prod_{i=0}^n \frac{1}{(x+i)}=A_k.$$
The general formula of $A_k$ is related to a binomial coefficient.
Once that the partial fraction decomposition is ready, the integration is straightforward.
