# Partial fractions of $\frac{1}{x^5-x^3}$

Is the following correct?

$$\frac{1}{(x^5-x^3)}=\frac{Ax+B}{(x^2-x)}+\frac{Cx+D}{(x^2-x)^2}+\frac{E}{x}$$

• Note that $(x^5-x^3)=x^3(x+1)(x-1)$ Commented Feb 16, 2016 at 21:39
• but $(x^5-x^3)=(x^2-x)^2x$
– gbox
Commented Feb 16, 2016 at 21:40
• $(x^2-x)^2=x^2(x^2-2x+1)=x^4-2x^3+x^2$... Commented Feb 16, 2016 at 21:41
• Do NOT make the mistake of saying that $(a+b)^2=a^2+b^2$!!! Commented Feb 16, 2016 at 21:43

No it is not, notice that:

$$\frac{1}{x^5-x^3}=\frac{1}{x^3(x-1)(x+1)}=\frac{A}{x^3}+\frac{B}{x^2}+\frac{C}{x-1}+\frac{D}{x}+\frac{E}{x+1}=\dots$$

Does $(x^2-x)^2x = (x^4-x^2)x = x^5-x^3$?

Check the first equality.

This one is better : $$\frac{1}{(x^5-x^3)}=\frac{a}{(x-1)}+\frac{cx^2+dx+e}{(x^3)}+\frac{b}{x+1}$$ using identification, we find $$a=b=1/2, c=-1, d=0, e=-1$$ then $$\frac{1}{(x^5-x^3)}=\frac{1}{2(x-1)}+-\frac{x^2+1}{x^3}+\frac{1}{2(x+1)}$$

• This is wrong , you need $\frac{1}{x}$ and $\frac{1}{x^2}$ too Commented Feb 16, 2016 at 21:45
• Fractions in the partialfraction decomposition with a degree $n$ polynomial for the denominator will have a degree $<n$ polynomial for the numerator. Factors which are repeated $n$ times will appear $n$ times as well, each of different powers. I.e. you should not have had an $Ax$ for the first or third fractions, and the middle fraction should instead have been $\frac{Cx^2+Dx+G}{x^3}+\frac{Hx+I}{x^2}+\frac{J}{x}$ Commented Feb 16, 2016 at 21:47
• I think that $Cx$ should be $Cx^2$. Here is why: quickmath.com/webMathematica3/quickmath/algebra/… Commented Feb 16, 2016 at 21:48
• Why above $x-1$ and $x+1$ it is not just A,B...? why Ax+B?
– gbox
Commented Feb 17, 2016 at 15:06