# What is the partial fraction of $\frac{x}{((x)^2+1)^2}$

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.

• 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 $\frac{x}{(x^{2} + 1)^{2}}$ – Adam Higgins Jan 22 '19 at 17:55
• 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. – Mark Bennet Jan 22 '19 at 17:56
• @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. – Mark Bennet Jan 22 '19 at 17:58
• @MarkBennet I'm not sure I understand your point – Adam Higgins Jan 22 '19 at 18:00
• @AdamHiggins You say this only works with linear factors. It can be done perfectly well with irreducible quadratic factors. – Mark Bennet Jan 22 '19 at 18:01

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 $$y=x^2+1$$.

• 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? – Naved THE Sheikh Jan 22 '19 at 18:13
• @NavedTHESheikh Definitely. – J.G. Jan 22 '19 at 18:18

Hint: Use that $$x^2+1=(x-i)(x+i)$$ where $$i^2=-1$$

• Thank you so much. I am yet trying it out. Will tell you if it works. 😉😊 – Naved THE Sheikh Jan 22 '19 at 18:14