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Question Number 58769 by Mr X pcx last updated on 29/Apr/19

$$decomposethefractionsinsideC\left(x\right)$$$$1)\frac{1}{{(x^2+1)}^3}$$$$2)\frac{1}{{(x^2+1)}^5}$$

Commented by maxmathsup by imad last updated on 01/May/19

$$1)F\left(x\right)=\frac{1}{{(x^2+1)}^3}=\frac{1}{{(x-i)}^3{(x+i)}^3}=\sum _{k=1}^3\frac{a_k}{{(x-i)}^k}+\sum _{k=1}^3\frac{b_k}{{(x+i)}^3}$$$$=\frac{a_1}{x-i}+\frac{a_1}{{(x-i)}^2}+\frac{a_3}{{(x-i)}^3}+\frac{b_1}{x+i}+\frac{b_2}{{(x+i)}^2}+\frac{b_3}{{(x+i)}^3}$$$$wehaveconj\left(F\left(x\right)\right)=F\left(x\right)\Rightarrow b_k={\stackrel{-}{a}}_{k}letdetermine{a}_k$$$$changementx-i=tgiveF\left(x\right)=G\left(t\right)=\frac{1}{t^3{(t+2i)}^3}letfind{D}_2\left(0\right)for$$$$w\left(t\right)=\frac{1}{{(t+2i)}^3}\Rightarrow w\left(t\right)=w\left(0\right)+\frac{t}{1!}{w}^{\left(1\right)}\left(0\right)+\frac{t^2}{2}{w}^{\left(2\right)}\left(0\right)+\frac{t^3}{3!}\xi \left(t\right)but$$$$w\left(t\right)={(t+2i)}^{-3}\Rightarrow w\left(0\right)={\left(2i\right)}^{-3}$$$$w^{{}^{\prime }}\left(t\right)=-3{(t+2i)}^{-4}\Rightarrow w^{\left(1\right)}\left(0\right)=-3{\left(2i\right)}^{-4}$$$$w^{\left(2\right)}\left(t\right)=12{(t+2i)}^{-5}\Rightarrow w^{\left(2\right)}\left(0\right)=12{\left(2i\right)}^{-5}\Rightarrow$$$$w\left(t\right)={\left(2i\right)}^{-3}-3{\left(2i\right)}^{-4}t+6{\left(2i\right)}^{-5}{t}^2+\frac{t^3}{3!}\xi \left(t\right)\Rightarrow \frac{w\left(t\right)}{t^3}$$$$=\frac{{\left(2i\right)}^{-3}}{t^3}-3{\left(2i\right)}^{-4}\frac{1}{t^2}+6{\left(2i\right)}^{-5}\frac{1}{t}+\frac{1}{3!}\xi \left(t\right)\Rightarrow$$$$F\left(x\right)=\frac{6{\left(2i\right)}^{-5}}{x-i}-\frac{3{\left(2i\right)}^{-4}}{{(x-i)}^2}+\frac{{\left(2i\right)}^{-3}}{{(x-i)}^3}+\frac{1}{3!}\xi (x-i)\Rightarrow$$$$a_1=\frac{6}{{\left(2i\right)}^5}=\frac{6}{2.16i}=\frac{3}{16i}=\frac{-3i}{16}$$$$a_2=\frac{-3}{{\left(2i\right)}^4}=\frac{-3}{16}and{a}_3=\frac{1}{{\left(2i\right)}^3}=\frac{1}{-8i}=\frac{i}{8}also{b}_1=\frac{3i}{16}$$$$b_2=\frac{-3}{16}and{b}_3=-\frac{i}{8}\Rightarrow$$$$F\left(x\right)=\frac{-3i}{16(x-i)}-\frac{3}{16{(x-i)}^2}+\frac{i}{8{(x-i)}^3}+\frac{3i}{16(x+i)}-\frac{3}{16{(x+i)}^2}-\frac{i}{8{(x+i)}^3}.$$$$$$

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