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Question Number 67564 by azizullah last updated on 28/Aug/19

Commented by mathmax by abdo last updated on 28/Aug/19

$$F\left(0\right)=0=1+1+2+f\Rightarrow f=-4\Rightarrow$$$$F\left(x\right)=-\frac{1}{x-1}+\frac{1}{x+1}+\frac{2}{x^2+1}-\frac{4}{{(x^2+1)}^2}.$$

Commented by mathmax by abdo last updated on 28/Aug/19

$$F\left(x\right)=\frac{8x^2}{(x^2+1)(1-x^4)}\Rightarrow F\left(x\right)=\frac{-8x^2}{(x^2+1)(x^2-1)(x^2+1)}\Rightarrow$$$$F\left(x\right)=\frac{-8x^2}{{(x^2+1)}^2(x-1)(x+1)}=\frac{a}{x-1}+\frac{b}{x+1}+\frac{cx+d}{x^2+1}+\frac{ex+f}{{(x^2+1)}^2}$$$$a={lim}_{x\to 1}(x-1)F\left(x\right)=\frac{-8}{2\times 4}=-1$$$$b={lim}_{x\to -1}(x+1)F\left(x\right)=\frac{-8}{4(-2)}=1\Rightarrow$$$$F\left(x\right)=\frac{-1}{x-1}+\frac{1}{x+1}+\frac{cx+d}{x^2+1}+\frac{ex+f}{{(x^2+1)}^2}$$$$F(-x)=F\left(x\right)\Rightarrow \frac{1}{x+1}-\frac{1}{x-1}+\frac{-cx+d}{x^2+1}+\frac{-ex+f}{{(x^2+1)}^2}\Rightarrow$$$$c=0ande=0\Rightarrow$$$$F\left(x\right)=\frac{-1}{x-1}+\frac{1}{x+1}+\frac{d}{x^2+1}+\frac{f}{{(x^2+1)}^2}$$$$=\frac{-2}{x^2-1}+\frac{d}{x^2+1}+\frac{f}{{(x^2+1)}^2}$$$${lim}_{x\to +\mathrm{\infty }}{x}^{2}F\left(x\right)=0=-2+d\Rightarrow d=2\Rightarrow$$$$F\left(x\right)=\frac{-1}{x-1}+\frac{1}{x+1}+\frac{2}{x^2+1}+\frac{f}{{(x^2+1)}^2}$$

Commented by Rasheed.Sindhi last updated on 29/Aug/19

$$SirprofAbdu$$$$Niceapproach,butI{can}^{\prime }tunderstand$$$$whyhaveyouputc=0\mathrm{\&}e=0.$$

Commented by mathmax by abdo last updated on 29/Aug/19

$$Fiseven\Rightarrow F(-x)=F\left(x\right)andthedecompositionisunique$$$$soweequalthecoefficient.$$

Commented by Rasheed.Sindhi last updated on 29/Aug/19

$$\mathcal{T}h\alpha n\mathcal{X}\mathcal{A}\mathcal{L}\mathrm{♡}\mathrm{\top }\int !⋎.$$

Commented by azizullah last updated on 29/Aug/19

$$\mathbf{Thanks}\mathbf{so}\mathbf{much}$$

Commented by Abdo msup. last updated on 30/Aug/19

$$youarewelcome$$

Commented by azizullah last updated on 13/Sep/19

$$\mathbf{excuse}\mathbf{me}\mathbf{sur}!\mathbf{i}\mathbf{need}\mathbf{all}\mathbf{theorems}\mathbf{proof}\mathbf{of}\mathbf{class}9\mathbf{th}.$$$$\mathbf{can}\mathbf{you}\mathbf{send}\mathbf{me}?\mathbf{i}\mathbf{share}\mathbf{it}\mathbf{in}\mathbf{our}\mathbf{note}.$$

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