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Question Number 34614 by math khazana by abdo last updated on 09/May/18

$$decomposeinsideR\left(x\right)thefraction$$$$F\left(x\right)=\frac{1}{{(x-3)}^6(x+2)}.$$

Commented by math khazana by abdo last updated on 09/May/18

$$changementx-3=tgive$$$$F\left(x\right)=g\left(t\right)=\frac{1}{t^6(t+5)}letfind{D}_5\left(0\right)for$$$$f\left(t\right)=\frac{1}{t+5}\Rightarrow f\left(t\right)=\sum _{k=0}^5\frac{f^{\left(k\right)}\left(0\right)}{k!}{t}^k+\frac{t^6}{6!}\xi \left(t\right)with$$$$\xi {\left(t\right)}_{t\to 0}\to 0wehave{f}^{\left(k\right)}\left(t\right)=\frac{{(-1)}^{k}k!}{{(t+5)}^{k+1}}\Rightarrow$$$$f^{\left(k\right)}\left(0\right)=\frac{{(-1)}^{k}k!}{5^{k+1}}\Rightarrow f\left(t\right)=\sum _{k=0}^5\frac{{(-1)}^k}{5^{k+1}}{t}^k+\frac{t^6}{6!}\xi \left(t\right)$$$$\Rightarrow g\left(t\right)=\sum _{k=0}^5\frac{{(-1)}^k}{5^{k+1}{t}^{6-k}}+\frac{1}{6!}\xi \left(t\right)$$$$={}_{6-k=p}\sum _{p=1}^6\frac{{(-1)}^{6-p}}{5^{7-p}{t}^p}+\frac{1}{6!}\xi \left(t\right)fromanother$$$$sideg\left(t\right)=\sum _{p=1}^6\frac{{\lambda }_p}{t^p}+\frac{a}{t+5}\Rightarrow {\lambda }_p=\frac{{(-1)}^{6-p}}{5^{7-p}}$$$$a={lim}_{t\to -5}(t+5)g\left(t\right)=\frac{1}{{(-5)}^6}\Rightarrow$$$$g\left(t\right)=\sum _{p=1}^6\frac{{(-1)}^{6-p}}{5^{7-p}{t}^p}+\frac{1}{5^6(t+5)}\Rightarrow$$$$F\left(x\right)=\sum _{p=1}^6\frac{{(-1)}^{6-\mathit{p}}}{5^{7-\mathit{p}}{(\mathit{x}-3)}^{\mathit{p}}}+\frac{1}{5^6(\mathit{x}+2)}.$$

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