# Partial fraction decomposition of $\frac{x+1}{x^2(x^3-1)}$

I am currently trying to find the partial fraction decomposition of

$$\frac{x+1}{x^2(x^3-1)}$$ But I am having some troubles: So far I have only simple fraction decompositions, where I could just factor the denominator and then compare coefficients and solve a system of linear equations. This method does not seem to work for this problem, however. I started out by saying that $$\frac{x+1}{x^2(x^3-1)} = \frac{x+1}{x^2(x^2+x+1)(x-1)}$$ Then I tried using the method I described above, so I did: $$\frac{x+1}{x^2(x^3-1)} =\frac{A}{x^2}+\frac{B}{x^2+x+1}+ \frac{C}{x-1}$$ I end up with a system of linear equations which does not have any solutions. As far as I have seen in various online calculators, I have to do something like this: $$\frac{x+1}{x^2(x^3-1)} =\frac{A}{x^2}+\frac Bx+\frac{Cx+D}{x^2+x+1}+ \frac{E}{x-1}$$ This genuinely confuses me. First of all, where does the $\frac Bx$ come from and why do I know that I need to write $\frac{Cx+D}{x^2+x+1}$ instead of just $\frac{C}{x^2+x+1}$? I know it has something to do with the degree of the polynomial in the denominator, but I would really appreciate any kind of help here.

The rule of thumbs is that on the numerator you always have to put a generic polynomial of degree one less the denominator. So if you have $x^2+x+1$, since it has degree two you have to put $Cx+D$. If you had $x^4+1$ you would need $Ax^3+Bx^2+Cx+D$. For repeated roots, say $(x-1)^3$ you would need $\frac {A}{x-1}$+$\frac {B}{(x-1)^2}$+ $\frac {C}{(x-1)^3}$

As a rule of thumb, $$\dfrac{p(x)}{q_1(x)^{m_1}q_2(x)^{m_2} \ldots q_1(x)^{m_n}} = \dfrac{r_{11}(x)}{q_1(x)} + \dfrac{r_{12}(x)}{q_1(x)^2} + \ldots \dfrac{r_{1m_1}(x)}{q_1(x)^{m_1}} + \dfrac{r_{21}(x)}{q_2(x)} + \ldots$$ where $m_i$ is the multiplicity of the terms in the denominator, and it's also important to note that the degree of $r_{ij}(x)$ is one less than the degree of $q_i(x)$

So in your case, the $\dfrac{B}{x} + \dfrac{A}{x^2}$ came from the $x^2$ in the denominator, since both $A$ and $B$ have degree strictly less than $x$.

And the $Cx + D$ has degree one less than $x^2 + x + 1$.

I hope that helps

• I think you mean $r_{ij}(x)$ is one less than the degree of $q_i(x)^j$. – mdave16 Mar 25 '17 at 15:35
• No i mean $q_i(x)$, an example would be the case above, $B$ is one less than the degree $x$ not $x^2$. – Yahya Fidouh Mar 25 '17 at 15:46

• when you have a multiple factor in the denominator, such as $x^2$, the corresponding simple elements happen up to the order of multiplicity, not at this order.
• Also, the numerator has degree less than the degree of the irreducible factor in the denominator (not the degree of the denominator in case of a multiple factor!)

So it should be $$\frac{x+1}{x^2(x^3-1)} =\frac{A'}{x}+\frac{A}{x^2}+\frac{B'x+B}{x^2+x+1}+ \frac{C}{x-1}\tag{1}$$

To determine the coefficients, you have smarter methods than reducing to the same denominator and solving the resulting linear system (5 equations in 5 unknowns here!).

Multiply both sides by $x^2(x^3-1)$ to remove all denominators and obtain: $$x+1=A'x(x^3-1)+A(x^3-1)+(B'x+B)x^2(x-1)+Cx^2(x^2+x+1) \tag2$$

• $x=0$: $(2)$ becomes $1=-A$, so $\;A=-1$.
• $x=1$: it becomes $2=3C$, so $\;C=\frac23$.
• $x=j=\mathrm e^{\tfrac{2i\pi}{3}}$: remembering $1+j+j^2=0$, you obtain \begin{align} j+1&=(B'j+B)j^2(j-1)=(B'j+B)(1-j^2)=B-B'+B'j-Bj^2\\&=2B-B'+(B+B')j \end{align} As $1$ and $j$ are linearly independent over $\mathbf R$, this yields the system $$\begin{cases} 2B-B'=1\\\phantom{2}B+B'=1 \end{cases}\iff B=\frac23,\;B'=\frac13$$ To determine the last coefficient, $A'$, multiply both sides of $(1)$: $$\frac{x+1}{x(x^3-1)} =A'+\frac{A}{x}+\frac{B'x+B}{x(x^2+x+1)}+ \frac{Cx}{x-1}$$ and let $x\to+\infty$, getting $$0=A'+C,\enspace \text{whence}\quad A'=-\frac23.$$ Finally, the decomposition is