Partial Fraction for $x^2/(x^2 +1)^2$

Question

$$\begin{aligned} \frac{\ x^2}{(x^2+1)^2} \\ \ \end{aligned}$$ I am new to partial fractions and this is what I have so far: $$\begin{aligned} \dfrac{(x^2)}{(x+1)^2} = \dfrac A{x-1}+\dfrac B{(x+1)^2} \ \end{aligned}$$

Am I on the right track? I was told it is impossible to take the partial fraction of this but that didn't seem right to me.

@NormalHuman I'd been told to not tag things with tags that are not associated with the question, so I kept it to partial fractions. — jake mckenzie, Oct 26 '15 at 23:11

score 2 · Accepted Answer · answered Oct 26 '15 at 23:03

2

Try $\frac{x^2}{(x^2+1)^2} =\frac{x^2+1-1}{(x^2+1)^2} =\frac1{x^2+1}-\frac{1}{(x^2+1)^2} $.

If you want to go into complex numbers, use $x^2+1 =(x+i)(x-i) $.

answered Oct 26 '15 at 23:03

marty cohen

107,799

score 0 · Answer 2 · answered Oct 26 '15 at 23:03

0

Consider writing \begin{align} \frac{x^2}{(x^2+1)^2}&=\frac{A}{x^2+1}+\frac{B}{(x^2+1)^2}; \end{align} you should be able to see what $A$ and $B$ should be.

answered Oct 26 '15 at 23:03

B. Pasternak

1,834

Partial Fraction for $x^2/(x^2 +1)^2$

2 Answers2