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

Question

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!

Answer 1

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 $y=x^2+1$.

Answer 2

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