Closed form for fixed $m$ to $\int\frac{dx}{x(x+1)(x+2)(x+3)...(x+m)}$ $I=\displaystyle\int\frac{dx}{x(x+1)(x+2)(x+3)...(x+m)}$
Attempt:
$\dfrac{ A_0 }{ x   }+\dfrac{ A_1 }{ x  +1 }+\dfrac{ A_2 }{ x +  2 }...+\dfrac{ A_m }{ x +m }  =\dfrac{1}{x(x+1)(x+2)(x+3)...(x+m)}$
But things got very messy.
I also thought that 1)applying integral by parts, or 2) taking terms one from left head, one from right hand and use some kind of a symmetry, or 3) using trigonometric identites etc.
I cannot see the solution, any hint, help would be perfect. Thank you in advanced.

$f(x) = \frac {1}{x(x+1)(x+2)\cdots(x+m)} = \frac {A_0}{x} + \cdots + \frac {A_n}{x+m}\\
\lim_\limits{x\to -n} (x-n)f(x) = A_n\\
A_n = \prod_\limits  {i\ne n} \frac 1{i-n}$
