Closed form for fixed $m$ to $\int\frac{dx}{x(x+1)(x+2)(x+3)...(x+m)}$

Starting from your attempt, notice that for $k \in \{0, \cdots, m\}$ we have

$$ A_k = \lim_{x\to-k} \frac{x+k}{x(x+1)\cdots(x+m)} = \frac{(-1)^k}{k!(m-k)!} = \frac{(-1)^k}{m!}\binom{m}{k}. $$

So we have

$$ \frac{1}{x(x+1)\cdots(x+m)} = \frac{1}{m!} \sum_{k=0}^{m} \binom{m}{k} \frac{(-1)^k}{x+k} $$

and now you can integrate term by term.