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Decompose $\frac{(x^3+x+1)}{(x^2+1)^2}$


Based on my understanding so far: The partials of the denominator are $(x^2+1)$ and $(x^2+1)^2$

$\frac{(x^3+x+1)}{(x^2+1)^2}$ can be decomposed into partial fractions as below:

$\frac{(x^3+x+1)}{(x^2+1)^2} = \frac{A}{x^2+1} +\frac{B}{(x^2+1)^2}$

multiply both sides by LCD which is $(x^2+1)^2$ we get:

$$x^3+x+1=A(x^2+1)+B$$ removing brackets

$$x^3+x+1=Ax^2+A+B$$

simplify

$$x(x^2+1)+1=A(x^2+1)+B$$

$\therefore$ looking at coefficients, $B=1$ and $A=x$.

Looking at the above, I don't seem to have a solution. Where am I going wrong? Am I on right path?

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  • $\begingroup$ Can you please put your comment into the original post? $\endgroup$ – user61527 Oct 1 '13 at 4:50
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By inspection, the partial fractions decomposition is $\dfrac{x}{x^2+1}+\dfrac{1}{(x^2+1)^2}$. This is because $x^3+x=x(x^2+1)$.

Remark: If the top were less simple, we would look for constants $A, B, C, D$ such that our expression is identically equal to $\frac{Ax+B}{x^2+1}+\frac{Cx+D}{(x^2+1)^2}$.

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  • $\begingroup$ I thought the numerator must always be some number? $\endgroup$ – Sylvester Oct 1 '13 at 4:59
  • $\begingroup$ When the denominator is a power of an irreducible quadratic, the numerator has shape $Px+Q$. Sometimes $P$ turns out to be $0$, but in general it does not. In your case, there do not exist $A$ and $B$ such that your expression is equal to $\frac{A}{x^2+1}+\frac{B}{(x^2+10^2}$. $\endgroup$ – André Nicolas Oct 1 '13 at 5:01
  • $\begingroup$ are you saying my workings are correct and that the values fors A and B are also correct and can be used in the decomposed partial fractions? I am not clear from your answer. $\endgroup$ – Sylvester Oct 1 '13 at 5:04
  • $\begingroup$ There are no constant values of $A$ and $B$ that work. You got lucky and doing it the wrong way happened to get the right answer. $\endgroup$ – André Nicolas Oct 1 '13 at 5:05
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    $\begingroup$ You can do it the long way, starting with $A,B,C,D$. In this case, there is an easier way, breaking up as $\frac{x^3+x}{(x^2+1)^2}+\frac{1}{(x^2+1)^2}$, and observing the cancellation in the first term, since $x^3+x=x(x^2+1)$. $\endgroup$ – André Nicolas Oct 1 '13 at 7:34

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