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I was trying to find the partial fraction of

$$\frac{x}{(x^2+1)^2}$$ By the method of assuming

$$\frac{x}{(x^2+1)^2}=\frac{(Ax+B)}{(x^2+1)} + \frac{(Cx+D)}{(x^2+1)^2} $$

But, my values for $A, B$ and $D$ are coming $0$. i.e. $$A=B=D=0$$ and $$C=1$$

Which is directly equal to $$\frac{x}{(x^2+1)^2}$$

So, technically I got NO solution or Partial Fraction

I am afraid if I'm doing any foolish mistake but, please help me out in this issue. I haven't practised Partial Fractions since a long time.

Thank you so much in advance! Great Day Ahead!

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    $\begingroup$ The ansatz that you are using only works when your rational function has a denominator that is a $\textbf{linear term}$ squared. The correct ansatz here is $\frac{x}{(x^{2} + 1)^{2}} = \frac{A}{x + i} + \frac{B}{x-i} + \frac{Cx + D}{(x+i)^{2}} + \frac{Ex + F}{(x - i)^{2}}$. I'm not sure if you wanted your partial fractions to have complex coefficients, but this will give you the complete partial fraction expansion of $\frac{x}{(x^{2} + 1)^{2}}$ $\endgroup$ – Adam Higgins Jan 22 at 17:55
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    $\begingroup$ This is right. You start off with a fraction in the form you require, and when you do the calculations they show that this is the right decomposition. You didn't get "no solution", just an unexpected solution. $\endgroup$ – Mark Bennet Jan 22 at 17:56
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    $\begingroup$ @AdamHiggins You can decompose into partial fractions in the way suggested in the question. For example if integrating a rational function over the reals the factorisation of the denominator into linear and quadratic factors with real coefficients does make sense. $\endgroup$ – Mark Bennet Jan 22 at 17:58
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    $\begingroup$ @MarkBennet I'm not sure I understand your point $\endgroup$ – Adam Higgins Jan 22 at 18:00
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    $\begingroup$ @AdamHiggins You say this only works with linear factors. It can be done perfectly well with irreducible quadratic factors. $\endgroup$ – Mark Bennet Jan 22 at 18:01
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Well, of course the solution of the form $\frac{Ax+B}{x^2+1}+\frac{Cx+D}{(x^2+1)^2}$ is $\frac{x}{(x^2+1)^2}$; that's what you started with. If you want another partial fractions expression, use complex numbers. If the point of the exercise is to integrate the function, you'd be better off substituting $y=x^2+1$.

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  • $\begingroup$ The cope of breaking $$\frac{x}{((x)^2+1)^2}$$ into Partial Fractions is to evaluate the value of it's Inverse Laplace Transform. Will it help, if I use complex numbers here? $\endgroup$ – Naved THE Sheikh Jan 22 at 18:13
  • $\begingroup$ @NavedTHESheikh Definitely. $\endgroup$ – J.G. Jan 22 at 18:18
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Hint: Use that $$x^2+1=(x-i)(x+i)$$ where $$i^2=-1$$

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  • $\begingroup$ Thank you so much. I am yet trying it out. Will tell you if it works. 😉😊 $\endgroup$ – Naved THE Sheikh Jan 22 at 18:14

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