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Question Number 94338 by mathmax by abdo last updated on 18/May/20

$$developpatintergrserief\left(x\right)=\frac{1}{(x+3)(x^2+4)}$$

Answered by mathmax by abdo last updated on 18/May/20

$$\mathrm{fist}\mathrm{let}\mathrm{find}{\mathrm{f}}^{\left(\mathrm{n}\right)}\left(0\right),\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{(\mathrm{x}+3)(\mathrm{x}-2\mathrm{i})(\mathrm{x}+2\mathrm{i})}=\frac{\mathrm{a}}{\mathrm{x}+3}+\frac{\mathrm{b}}{\mathrm{x}-2\mathrm{i}}+\frac{\mathrm{c}}{\mathrm{x}+2\mathrm{i}}$$$$\mathrm{a}=\frac{1}{13}\mathrm{b}=\frac{1}{(2\mathrm{i}+3)4\mathrm{i}}\mathrm{and}\mathrm{c}=\frac{1}{-4\mathrm{i}(-2\mathrm{i}+3)}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{13(\mathrm{x}+3)}+\frac{1}{\mathrm{i}(8\mathrm{i}+12)(\mathrm{x}-2\mathrm{i})}+\frac{1}{4\mathrm{i}(2\mathrm{i}-3)(\mathrm{x}+2\mathrm{i})}\Rightarrow$$$${\mathrm{f}}^{\left(\mathrm{n}\right)}\left(\mathrm{x}\right)=\frac{1}{13}\times \frac{{(-1)}^{\mathrm{n}}\mathrm{n}!}{{(\mathrm{x}+3)}^{\mathrm{n}+1}}-\frac{\mathrm{i}}{8\mathrm{i}+12}\times \frac{{(-1)}^{\mathrm{n}}\mathrm{n}!}{{(\mathrm{x}-2\mathrm{i})}^{\mathrm{n}+1}}-\frac{\mathrm{i}{(-1)}^{\mathrm{n}}\mathrm{n}!}{(8\mathrm{i}-12){(\mathrm{x}+2\mathrm{i})}^{\mathrm{n}+1}}$$$$\Rightarrow {\mathrm{f}}^{\left(\mathrm{n}\right)}\left(0\right)=\frac{1}{13}\times \frac{{(-1)}^{\mathrm{n}}\mathrm{n}!}{3^{\mathrm{n}+1}}-\frac{\mathrm{i}}{8\mathrm{i}+12}\frac{{(-1)}^{\mathrm{n}}\mathrm{n}!}{{(-2\mathrm{i})}^{\mathrm{n}+1}}-\frac{\mathrm{i}{(-1)}^{\mathrm{n}}\mathrm{n}!}{(8\mathrm{i}-12){\left(2\mathrm{i}\right)}^{\mathrm{n}+1}}$$$$\mathrm{f}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\mathrm{f}}^{\left(\mathrm{n}\right)}\left(0\right)}{\mathrm{n}!}{\mathrm{x}}^{\mathrm{n}}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{1}{13}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{3^{\mathrm{n}+1}}{\mathrm{x}}^{\mathrm{n}}$$$$-\frac{\mathrm{i}}{8\mathrm{i}+12}\mathrm{\Sigma }\frac{{(-1)}^{\mathrm{n}}}{{(-2\mathrm{i})}^{\mathrm{n}+1}}{\mathrm{x}}^{\mathrm{n}}-\frac{\mathrm{i}}{8\mathrm{i}-12}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{\left(2\mathrm{i}\right)}^{\mathrm{n}+1}}{\mathrm{x}}^{\mathrm{n}}$$$$$$

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