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Question Number 83063 by M±th+et£s last updated on 27/Feb/20

Commented by M±th+et£s last updated on 27/Feb/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Commented by abdomathmax last updated on 27/Feb/20

let decompose F(x)=(1/(x(x+1)....(x+m))) =(1/(Π_(k=0) ^m (x+k))) ⇒F(x)=Σ_(k=0) ^m (a_k /(x+k)) a_k =lim_(x→−k) (x+k)F(x) we hsve F(x)=(1/(x(x+1)...(x+k−1)(x+k)(x+k+1)...(x+m))) ⇒a_k =(1/((−k)(−k+1)....(−k+k−1)(−k+k+1)...(−k+m))) =(1/((−1)^k k!(m−k)!)) =(((−1)^k )/(k!(m−k)!)) ⇒ F(x)=Σ_(k=0) ^m (((−1)^k )/(k!(m−k)!(x+k))) ⇒ ∫ F(x)dx =Σ_(k=0) ^m (((−1)^k )/(k!(m−k)!))ln∣x+k∣ +C

$${let}\:{decompose}\:{F}\left({x}\right)=\frac{\mathrm{1}}{{x}\left({x}+\mathrm{1}\right)....\left({x}+{m}\right)} \\ $$$$=\frac{\mathrm{1}}{\prod_{{k}=\mathrm{0}} ^{{m}} \left({x}+{k}\right)}\:\Rightarrow{F}\left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{m}} \:\frac{{a}_{{k}} }{{x}+{k}} \\ $$$${a}_{{k}} ={lim}_{{x}\rightarrow−{k}} \:\left({x}+{k}\right){F}\left({x}\right) \\ $$$${we}\:{hsve}\:{F}\left({x}\right)=\frac{\mathrm{1}}{{x}\left({x}+\mathrm{1}\right)...\left({x}+{k}−\mathrm{1}\right)\left({x}+{k}\right)\left({x}+{k}+\mathrm{1}\right)...\left({x}+{m}\right)} \\ $$$$\Rightarrow{a}_{{k}} =\frac{\mathrm{1}}{\left(−{k}\right)\left(−{k}+\mathrm{1}\right)....\left(−{k}+{k}−\mathrm{1}\right)\left(−{k}+{k}+\mathrm{1}\right)...\left(−{k}+{m}\right)} \\ $$$$=\frac{\mathrm{1}}{\left(−\mathrm{1}\right)^{{k}} {k}!\left({m}−{k}\right)!}\:=\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}!\left({m}−{k}\right)!}\:\Rightarrow \\ $$$${F}\left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{m}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}!\left({m}−{k}\right)!\left({x}+{k}\right)}\:\Rightarrow \\ $$$$\int\:{F}\left({x}\right){dx}\:=\sum_{{k}=\mathrm{0}} ^{{m}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}!\left({m}−{k}\right)!}{ln}\mid{x}+{k}\mid\:+{C} \\ $$

Commented by mathmax by abdo last updated on 27/Feb/20

you are welcome

$${you}\:{are}\:{welcome} \\ $$

Answered by mind is power last updated on 27/Feb/20

(1/(Π_(k=0) ^m (x+k)))Σ_(j=0) ^m (a_j /(x+j)) a_j =(1/(Π_(k=0,k≠j) ^m (k−j)))=(1/(Π_(k=0) ^(j−1) (k−j).Π_(j+1) ^m (k−j)))=(((−1)^j )/(j!.(m−j)!))=a_j =Σ_(j=0) ^m (((−1)^j )/(j!(m−j)!)).(1/(x+j)) ∫(dx/(Π_(k=0) ^m (x+k)))=∫Σ_(j=0) ^m (((−1)^j )/(j!(m−j)!)).(1/(x+j))dx=Σ_(j=0) ^m (((−1)^j )/(j!(m−j)!))∫(dx/(x+j)) =Σ_(j=0) ^m (((−1)^j )/(j!(m−j)!))ln(∣x+j∣)+c

$$\frac{\mathrm{1}}{\underset{{k}=\mathrm{0}} {\overset{{m}} {\prod}}\left({x}+{k}\right)}\underset{{j}=\mathrm{0}} {\overset{{m}} {\sum}}\frac{{a}_{{j}} }{{x}+{j}} \\ $$$${a}_{{j}} =\frac{\mathrm{1}}{\underset{{k}=\mathrm{0},{k}\neq{j}} {\overset{{m}} {\prod}}\left({k}−{j}\right)}=\frac{\mathrm{1}}{\underset{{k}=\mathrm{0}} {\overset{{j}−\mathrm{1}} {\prod}}\left({k}−{j}\right).\underset{{j}+\mathrm{1}} {\overset{{m}} {\prod}}\left({k}−{j}\right)}=\frac{\left(−\mathrm{1}\right)^{{j}} }{{j}!.\left({m}−{j}\right)!}={a}_{{j}} \\ $$$$=\underset{\mathrm{j}=\mathrm{0}} {\overset{\mathrm{m}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{j}} }{{j}!\left({m}−{j}\right)!}.\frac{\mathrm{1}}{{x}+\mathrm{j}} \\ $$$$\int\frac{{dx}}{\underset{{k}=\mathrm{0}} {\overset{{m}} {\prod}}\left({x}+{k}\right)}=\int\underset{{j}=\mathrm{0}} {\overset{{m}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{j}} }{{j}!\left({m}−{j}\right)!}.\frac{\mathrm{1}}{{x}+{j}}{dx}=\underset{{j}=\mathrm{0}} {\overset{{m}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{j}} }{{j}!\left({m}−{j}\right)!}\int\frac{{dx}}{{x}+{j}} \\ $$$$=\underset{{j}=\mathrm{0}} {\overset{{m}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{j}} }{{j}!\left({m}−{j}\right)!}{ln}\left(\mid{x}+{j}\mid\right)+\mathrm{c} \\ $$

Commented by M±th+et£s last updated on 27/Feb/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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