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Question-146577

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Question Number 146577 by bekzodjumayev last updated on 14/Jul/21
Commented by bekzodjumayev last updated on 14/Jul/21
Help
$${Help} \\ $$$$ \\ $$
Answered by Ar Brandon last updated on 14/Jul/21
I=∫(dx/(x(x+1)(x+2)∙∙∙(x+m))) =∫((1/(m!))∙(1/x)−(1/((m−1)!))∙(1/(x+1))+(1/(2(m−2)!))∙(1/(x+2)) (−(1/(3!(m−3)!))∙(1/(x+3))+∙∙∙+(−1)^m (1/(m!))∙(1/(x+m)))dx =∫Σ_(k=0) ^m (−1)^k (1/(k!(m−k)!))∙(1/(x+k))dx=Σ_(k=0) ^m (−1)^k ((ln∣x+k∣)/(k!(m−k)!))+C
$$\mathrm{I}=\int\frac{\mathrm{dx}}{\mathrm{x}\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}+\mathrm{2}\right)\centerdot\centerdot\centerdot\left(\mathrm{x}+\mathrm{m}\right)} \\ $$$$\:\:=\int\left(\frac{\mathrm{1}}{\mathrm{m}!}\centerdot\frac{\mathrm{1}}{\mathrm{x}}−\frac{\mathrm{1}}{\left(\mathrm{m}−\mathrm{1}\right)!}\centerdot\frac{\mathrm{1}}{\mathrm{x}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{m}−\mathrm{2}\right)!}\centerdot\frac{\mathrm{1}}{\mathrm{x}+\mathrm{2}}\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\left(−\frac{\mathrm{1}}{\mathrm{3}!\left(\mathrm{m}−\mathrm{3}\right)!}\centerdot\frac{\mathrm{1}}{\mathrm{x}+\mathrm{3}}+\centerdot\centerdot\centerdot+\left(−\mathrm{1}\right)^{\mathrm{m}} \frac{\mathrm{1}}{\mathrm{m}!}\centerdot\frac{\mathrm{1}}{\mathrm{x}+\mathrm{m}}\right)\mathrm{dx} \\ $$$$\:\:=\int\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{m}} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{k}} \frac{\mathrm{1}}{\mathrm{k}!\left(\mathrm{m}−\mathrm{k}\right)!}\centerdot\frac{\mathrm{1}}{\mathrm{x}+\mathrm{k}}\mathrm{dx}=\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{m}} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{k}} \frac{\mathrm{ln}\mid\mathrm{x}+\mathrm{k}\mid}{\mathrm{k}!\left(\mathrm{m}−\mathrm{k}\right)!}+\mathrm{C} \\ $$
Commented by bekzodjumayev last updated on 14/Jul/21
Thank you
$${Thank}\:{you} \\ $$
Answered by mathmax by abdo last updated on 14/Jul/21
Ψ=∫ (dx/(x(x+1)(x+2).....(x+m))) let decompose F(x)=(1/(x(x+1)....(x+m)))=Σ_(k=0) ^m (a_k /(x+k)) a_k =lim_(x→−k) (x+k)F(x) =lim_(x→−k) (x+k)×(1/(x(x+1).....(x+k−1)(x+k)(x+k+1)....(x+m))) =(1/((−k)(−k+1).....(−1)×1.2.....(−k+m))) =(1/((−1)^k k!(m−k)!))=(((−1)^k )/(m!))×((m!)/(k!(m−k)!))=(((−1)^k )/(m!))C_m ^k ⇒ F(x)=(1/(m!))Σ_(k=0) ^m (((−1)^k C_m ^k )/(x+k)) ⇒ Ψ=∫ F(x)dx=(1/(m!))Σ_(k=0) ^m (−1)^k C_m ^k log∣x+k∣ +C
$$\Psi=\int\:\:\frac{\mathrm{dx}}{\mathrm{x}\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}+\mathrm{2}\right)…..\left(\mathrm{x}+\mathrm{m}\right)} \\ $$$$\mathrm{let}\:\mathrm{decompose}\:\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{x}\left(\mathrm{x}+\mathrm{1}\right)….\left(\mathrm{x}+\mathrm{m}\right)}=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{m}} \:\frac{\mathrm{a}_{\mathrm{k}} }{\mathrm{x}+\mathrm{k}} \\ $$$$\mathrm{a}_{\mathrm{k}} =\mathrm{lim}_{\mathrm{x}\rightarrow−\mathrm{k}} \left(\mathrm{x}+\mathrm{k}\right)\mathrm{F}\left(\mathrm{x}\right) \\ $$$$=\mathrm{lim}_{\mathrm{x}\rightarrow−\mathrm{k}} \:\:\:\left(\mathrm{x}+\mathrm{k}\right)×\frac{\mathrm{1}}{\mathrm{x}\left(\mathrm{x}+\mathrm{1}\right)…..\left(\mathrm{x}+\mathrm{k}−\mathrm{1}\right)\left(\mathrm{x}+\mathrm{k}\right)\left(\mathrm{x}+\mathrm{k}+\mathrm{1}\right)….\left(\mathrm{x}+\mathrm{m}\right)} \\ $$$$=\frac{\mathrm{1}}{\left(−\mathrm{k}\right)\left(−\mathrm{k}+\mathrm{1}\right)…..\left(−\mathrm{1}\right)×\mathrm{1}.\mathrm{2}…..\left(−\mathrm{k}+\mathrm{m}\right)} \\ $$$$=\frac{\mathrm{1}}{\left(−\mathrm{1}\right)^{\mathrm{k}} \mathrm{k}!\left(\mathrm{m}−\mathrm{k}\right)!}=\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{m}!}×\frac{\mathrm{m}!}{\mathrm{k}!\left(\mathrm{m}−\mathrm{k}\right)!}=\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{m}!}\mathrm{C}_{\mathrm{m}} ^{\mathrm{k}} \:\Rightarrow \\ $$$$\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{m}!}\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{m}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} \:\mathrm{C}_{\mathrm{m}} ^{\mathrm{k}} }{\mathrm{x}+\mathrm{k}}\:\Rightarrow \\ $$$$\Psi=\int\:\mathrm{F}\left(\mathrm{x}\right)\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{m}!}\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{m}} \:\left(−\mathrm{1}\right)^{\mathrm{k}} \:\mathrm{C}_{\mathrm{m}} ^{\mathrm{k}} \mathrm{log}\mid\mathrm{x}+\mathrm{k}\mid\:+\mathrm{C} \\ $$$$ \\ $$

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