<s>[INST] 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.
 [/INST] Maybe the following reduction will be helpful:
$$x(x+1)(x+2)...(x+(m-2))(x+(m-1))(x+m)=x(x+m)(x+1)(x+(m-1))(x+2)(x+m(m-1))...=(x^2+mx)(x^2+mx+1(m-1))(x^2+mx+2(m-2))$$
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