Question Number 71831 by ahmadshahhimat775@gmail.com last updated on 20/Oct/19
Commented by kaivan.ahmadi last updated on 20/Oct/19
$${hi}\:{mr}\:{ahmadi} \\ $$$${where}\:{are}\:{you}\:{from}? \\ $$
Commented by Abdo msup. last updated on 21/Oct/19
$${let}\:{decompose}\:{F}\left({x}\right)=\frac{\mathrm{1}}{\prod_{{k}=\mathrm{0}} ^{{m}} \left({x}+{k}\right)} \\ $$$${F}\left({x}\right)\:=\sum_{{k}=\mathrm{0}} ^{{m}} \:\frac{{a}_{{k}} }{{x}+{k}} \\ $$$${a}_{{k}} ={lim}_{{x}\rightarrow−{k}} \:\:\left({x}+{k}\right){F}\left({x}\right) \\ $$$$={lim}_{{x}\rightarrow−{k}} \:\:\:\frac{\mathrm{1}}{{x}\left({x}+\mathrm{1}\right)…\left({x}+{k}−\mathrm{1}\right)\left({x}+{k}+\mathrm{1}\right)…\left({x}+{m}\right)} \\ $$$$=\frac{\mathrm{1}}{\left(−{k}\right)\left(−{k}+\mathrm{1}\right)….\left(−{k}+{k}−\mathrm{1}\right)\left(−{k}+{k}+\mathrm{1}\right)…\left(−{k}+{m}\right)} \\ $$$$=\frac{\mathrm{1}}{\left(−\mathrm{1}\right)^{{k}} {k}!\left({m}−{k}\right)!}\:=\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}!\left({m}−{k}\right)!}\:=\frac{\mathrm{1}}{{m}!}\left(−\mathrm{1}\right)^{{k}} \:{C}_{{m}} ^{{k}} \:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{\mathrm{1}}{{m}!}\sum_{{k}=\mathrm{0}} ^{{m}} \:\frac{\left(−\mathrm{1}\right)^{{k}} \:{C}_{{m}} ^{{k}} }{{x}+{k}}\:\Rightarrow \\ $$$$\int\:\:\:\:\frac{{dx}}{{x}\left({x}+\mathrm{1}\right)….\left({x}+{m}\right)}\:=\frac{\mathrm{1}}{{m}!}\sum_{{k}=\mathrm{0}} ^{{m}} \left(−\mathrm{1}\right)^{{k}} \:{C}_{{m}} ^{{k}} {ln}\mid{x}+{k}\mid\:+{C} \\ $$$$ \\ $$$$ \\ $$
Answered by MJS last updated on 20/Oct/19
$$=\underset{{n}=\mathrm{0}} {\overset{{m}} {\sum}}\left[\left({m}−{n}\right)!{n}!\mathrm{cos}\:\left({n}\pi\right)\:×\mathrm{ln}\:\left({x}+{n}\right)\right]+{C} \\ $$