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I want to do partial fraction decomposition on the following rational function:


So I proceed as follows:

$$\begin{align} \frac{1}{x^2(1+x^2)^3} &= \frac{A}{x} + \frac{B}{x^2} + \frac{Cx + D}{1 + x^2} + \frac{Ex + F}{(1 + x^2)^2} + \frac{Gx + H}{(1 + x^2)^3} \\ 1 = &Ax(1+x^2)^3 + B(1+x^2)^3 + (Cx + D)x^2(1+x^2)^2 \\ & + (Ex + F)x^2(1+x^2) + (Gx + H)(x^2) \\ 1 = &(A+C)x^7 + (B+D)x^6 + (3A+2C+E)x^5 + (3B+2D)x^4 \\ & + (3A+C+E+G)x^3 + (3B+D+H)x^2 +(A)x + (B)\\ \end{align}$$

But from this, $B=1$, $B=-D$, and $3B+2D=0$, which is an inconsistent system. What am I doing wrong?

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The Latex looks really bad, does anyone know how I can fix it? – Ataraxia Jun 20 '13 at 1:02
Since it can be written as a function of $x^2$, you can be sure that $A=0$ - basically, you want to come up with the partial fraction expansion for $\dfrac 1{u(1+u)^3}$. Then replace $u$ with $x^2$. So you know that $A=C=E=G=0$. – Thomas Andrews Jun 20 '13 at 1:03
It looks bad because you have long equations. I broke it up into 2 lines. – Calvin Lin Jun 20 '13 at 1:05
$ \frac{1}{x^2(1+x^2)^3} = \frac{1}{x^2} - \frac{1}{1+x^2} - \frac{1}{(1+x^2)^2} - \frac{1}{(1+x^2)^3}$. – Calvin Lin Jun 20 '13 at 1:07
up vote 3 down vote accepted

Your final expression is almost correct; you're just forgetting the $F$. The coefficient of $x^4$ should be: $$ 3B + 2D + F $$ and the coefficient of $x^2$ should be: $$ 3B + D + F + H $$

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Not answering your question, but I see $x^2$ everywhere, so it is better to find a partial fractions expression for $\dfrac{1}{u(1+u)^3}$.

So we are looking for $A$, $B$, $C$, $D$ such that $$\frac{1}{u(1+u)^3}=\frac{A}{u}+\frac{B}{1+u}+\frac{C}{(1+u)^2}+\frac{D}{(1+u)^3}.$$ When we bring the right-hand side to a common denominator $u(1+u)^3$, the numerator becomes $$A(1+u)^3+Bu(1+u)^2+Cu(1+u)+Du.$$ This is identically equal to $1$. Put $u=0$. We get $A=1$. Put $u=-1$. We get $D=-1$. The coefficient of $u^3$ is $A+B$ so $B=-1$. The coefficient of $u^2$ is $3A+2B+C$, so $C=-1$.

Now replace $u$ by $x^2$.

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