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Question Number 71831 by ahmadshahhimat775@gmail.com last updated on 20/Oct/19

Commented by kaivan.ahmadi last updated on 20/Oct/19

$$himrahmadi$$$$whereareyoufrom?$$

Commented by Abdo msup. last updated on 21/Oct/19

$$letdecomposeF\left(x\right)=\frac{1}{\prod _{k=0}^m(x+k)}$$$$F\left(x\right)=\sum _{k=0}^m\frac{a_k}{x+k}$$$$a_k={lim}_{x\to -k}(x+k)F\left(x\right)$$$$={lim}_{x\to -k}\frac{1}{x(x+1)...(x+k-1)(x+k+1)...(x+m)}$$$$=\frac{1}{(-k)(-k+1)....(-k+k-1)(-k+k+1)...(-k+m)}$$$$=\frac{1}{{(-1)}^{k}k!(m-k)!}=\frac{{(-1)}^k}{k!(m-k)!}=\frac{1}{m!}{(-1)}^k{C}_m^k\Rightarrow$$$$F\left(x\right)=\frac{1}{m!}\sum _{k=0}^m\frac{{(-1)}^k{C}_m^k}{x+k}\Rightarrow$$$$\int \frac{dx}{x(x+1)....(x+m)}=\frac{1}{m!}\sum _{k=0}^m{(-1)}^k{C}_m^{k}ln\mid x+k\mid +C$$$$$$$$$$

Answered by MJS last updated on 20/Oct/19

$$=\underset{n=0}{\sum ^m}[(m-n)!n!\cos \left(n\pi \right)\times \ln (x+n)]+C$$

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