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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

$$thankyousir$$

Commented by abdomathmax last updated on 27/Feb/20

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

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

$$youarewelcome$$

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

$$\frac{1}{\underset{k=0}{\prod ^m}(x+k)}\underset{j=0}{\sum ^m}\frac{a_j}{x+j}$$$$a_j=\frac{1}{\underset{k=0,k\ne j}{\prod ^m}(k-j)}=\frac{1}{\underset{k=0}{\prod ^{j-1}}(k-j).\underset{j+1}{\prod ^m}(k-j)}=\frac{{(-1)}^j}{j!.(m-j)!}=a_j$$$$=\underset{\mathrm{j}=0}{\sum ^{\mathrm{m}}}\frac{{(-1)}^j}{j!(m-j)!}.\frac{1}{x+\mathrm{j}}$$$$\int \frac{dx}{\underset{k=0}{\prod ^m}(x+k)}=\int \underset{j=0}{\sum ^m}\frac{{(-1)}^j}{j!(m-j)!}.\frac{1}{x+j}dx=\underset{j=0}{\sum ^m}\frac{{(-1)}^j}{j!(m-j)!}\int \frac{dx}{x+j}$$$$=\underset{j=0}{\sum ^m}\frac{{(-1)}^j}{j!(m-j)!}ln(\mid x+j\mid )+\mathrm{c}$$

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

$$thankyousir$$

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