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.

You may use:
$$\prod_{k=0}^{n} \frac{1}{x+k}=\frac{1}{n!}\sum_{k=1}^{n} (-1)^{k-1} \frac{k{n \choose k}}{x(x+k)}=\frac{1}{n!}\sum_{k=1}^{n}(-1)^{k-1} {n \choose k} [\frac{1}{x}-\frac{1}{x+k}]$$
So $$\int \prod_{k=0}^{n} \frac{1}{x+k} dx=\frac{1}{n!}\sum_{k=1}^{n}(-1)^{k-1} {n \choose k}[\ln x- \ln (x+k)]$$
