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Express $$\frac{2x}{(x^2 + 1)(x + 1)^2} = \frac{A_1 x + A_2}{(x^2+1)} + \frac{B}{(x+1)^2} + \frac{C}{x+1}$$ in partial fractions. I know I have to decompose it into three fractions with numerators $(x^2 + 1), (x + 1)$ and $(x + 1)^2$.

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As you know, $$\frac{2x}{(x^2+1)(x+1)^2}=\frac A{x+1}+\frac B{(x+1)^2}+\frac{Cx+D}{x^2+1}$$

Multiply either sides by $(x^2+1)(x+1)^2$ to get $$2x=A(x+1)(x^2+1)+B(x^2+1)+(Cx+D)(x+1)^2$$

Now arrange the Right Hand side as the descending power of $x$

and compare the coefficients of the different powers of $x$

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  • $\begingroup$ I've got A=0, B=-1, C=0 and D=1. Can you verify whether this is correct or not? Would be very much appreciated... $\endgroup$ – Vladimir Nabokov Oct 13 '13 at 19:01
  • $\begingroup$ @LeonLeibovici,It seems correct to me. $\endgroup$ – lab bhattacharjee Oct 13 '13 at 19:09
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Let denote your fraction by $F(x)$ hence we have $$B=(x+1)^2F(x)\big|_{x=-1}=-1$$ $$A_1i+A_2=(x^2+1)F(x)\big|_{x=i}=\frac{2i}{(1+i)^2}=1\quad \text{so}\quad A_1=0\ ;\ A_2=1$$ and finally we have $$\lim_{x\to\infty}xF(x)=0=A_1+C\quad \text{so}\quad C=-A_1=0$$

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  • $\begingroup$ Thanks for the support, dear friend! $\endgroup$ – Namaste Mar 13 '14 at 16:47

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