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Question Number 32333 by abdo imad last updated on 23/Mar/18

$$decomposeF\left(x\right)=\frac{1}{{(1-x)}^2(1-x^2)}insideR\left(x\right).$$

Commented by abdo imad last updated on 01/Apr/18

$$F\left(x\right)=\frac{1}{{(1-x)}^2(1-x)(1+x)}=\frac{1}{{(1-x)}^3(1+x)}=\frac{-1}{{(x-1)}^3(x+1)}$$$$=\frac{a}{x-1}+\frac{b}{{(x-1)}^2}+\frac{c}{{(x-1)}^3}+\frac{d}{x+1}$$$$c={lim}_{x\to 1}{(x-1)}^{3}F\left(x\right)=\frac{-1}{2}$$$$d={lim}_{x\to -1}(x+1)F\left(x\right)=\frac{1}{8}\Rightarrow$$$$F\left(x\right)=\frac{a}{x-1}+\frac{b}{{(x-1)}^2}-\frac{1}{2{(x-1)}^3}+\frac{1}{8(x+1)}$$$$F\left(0\right)=1=-a+b+\frac{1}{2}+\frac{1}{8}=-a+b+\frac{5}{8}$$$$\Rightarrow -a+b=1-\frac{5}{8}=\frac{3}{8}$$$$F\left(2\right)=\frac{-1}{3}=a+b-\frac{1}{2}+\frac{1}{24}=a+b-\frac{11}{24}$$$$\Rightarrow a+b=-\frac{1}{3}+\frac{11}{24}=\frac{3}{24}=\frac{1}{8}wegetthesystem$$$$-a+b=\frac{3}{8}anda+b=\frac{1}{8}\Rightarrow 2b=\frac{1}{2}\Rightarrow b=\frac{1}{4}$$$$a=\frac{1}{8}-b=\frac{1}{8}-\frac{1}{4}=-\frac{1}{8}and$$$$F\left(x\right)=\frac{-1}{8(x-1)}+\frac{1}{4{(x-1)}^2}-\frac{1}{2{(x-1)}^3}+\frac{1}{8(x+1)}.$$

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