Integration of $\frac{1}{x(x+1)(x+2)...(x+m)}$ I came across this question,
$$\int \frac{1}{x(x+1)(x+2)...(x+m)} dx$$
I tried to split the rational function into partial fractions
$$ \frac{A}{x}+\frac{B}{x+1}+\frac{C}{x+2}+...+\frac{Z}{x+m}$$
Not really sure how to proceed from here.
Can someone enlighten me with step by step explanations? I kinda get confused easily when certain steps are skipped.

Note
$$\frac{1}{x(x+1)(x+2)...(x+m)}=\sum_{k =0}^{m} \frac{a_k}{x+k}
$$
where $a_k$ are obtained as follows
\begin{align}
a_k &=\lim_{x\to -k} \frac{x+k}{x(x+1)(x+2)...(x+k)...(x+m)}\\
&= \frac{1}{[(-k)(1-k)(2-k)(-2)(-1)]\cdot[(1)(2)...(m-k-1)(m-k)]}\\
&=\frac1{(-1)^k k!(m-k)!}
\end{align}
Thus
$$\int \frac{dx}{x(x+1)(x+2)...(x+m)}
=\int \sum_{k =0}^{m} \frac{a_k}{x+k}dx
= \sum_{k =0}^{m} \frac{(-1)^k\ln|x+k|}{k!(m-k)!}+C
$$
