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Question Number 139476 by aliibrahim1 last updated on 27/Apr/21

Answered by qaz last updated on 27/Apr/21

$$I=\int \frac{dx}{x(x+1)(x+2)...(x+n)}$$$$f\left(x\right)=\frac{1}{x(x+1)(x+2)...(x+n)}=\frac{A_0}{x}+\frac{A_1}{x+1}+\frac{A_2}{x+2}+...+\frac{A_n}{x+n}$$$$A_0=\underset{x\to 0}{\lim }xf\left(x\right)=\frac{1}{n!}$$$$A_1=\underset{x\to -1}{\lim }(x+1)f\left(x\right)=\frac{1}{(-1)(n-1)!}$$$$A_2=\underset{x\to -2}{\lim }(x+2)f\left(x\right)=\frac{1}{(-2)(-1)(n-2)!}$$$$....$$$$A_n=\underset{x\to -n}{\lim }(x+n)f\left(x\right)=\frac{1}{(-n)(-n+1)(-n+2)\cdot ...\cdot 2\cdot 1}$$$$I=\frac{ln\mid x\mid }{n!}-\frac{ln\mid x+1\mid }{(n-1)!}+\frac{ln\mid x+2\mid }{2!(n-2)!}-..+{(-1)}^k\cdot \frac{ln\mid x+k\mid }{k!(n-k)!}+...+{(-1)}^n\frac{ln\mid x+n\mid }{n!}C$$$$=\underset{k=0}{\sum ^n}{(-1)}^k\frac{ln\mid x+k\mid }{k!(n-k)!}+C$$

Commented by aliibrahim1 last updated on 27/Apr/21

$$thxsir$$

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