Partial fraction expansion question

Question

I have to integrate following expression (but integration is not the problem): $$\frac{x^2+3x-2}{(x-1)(x^2+x+1)^2}$$

It is pretty obvious that: $$\frac{x^2+3x-2}{(x-1)(x^2+x+1)^2}=\frac{A}{x-1} + \frac{Mx+N}{x^2+x+1} + \frac{Px+Q}{(x^2+x+1)^2}$$

The first and the easiest step is to find an $A$: $$A=\frac{x^2+3x-2}{(x^2+x+1)^2}, x=1$$ $$A=\frac{2}{9}$$

And then there comes a problem - I don't know how to do the rest. I tried to multiply the whole thing by $(x^2+x+1)^2$ and differentiate, but it didn't seem to be useful at all, especially because $(x^2+x+1)^2$ doesn't have real roots.

As popping900 suggested. I can take just four different x values and solve system of for equations, but i would like to see a more elegant or shorter solution, if one exists

One way is to substitute in 4 different values of x, to solve a system of four equations and four unknowns. It might sound troublesome, however, based on the choices of x, it might not be too bad. e.g. take x=0,x=-1,x=2,x=-2. — popping900, Aug 10 '21 at 13:09
Note that $A(x^2 + x + 1)^2 = Ax^4 +\ ...\ $, clearly $A$ must be $0$. — jameselmore, Aug 10 '21 at 13:12
I get $A =0,\ M = 0,\ N = 1,\ P = 2,\ Q = -3$. When I multiply these out I get the original fraction... — jameselmore, Aug 10 '21 at 13:18
The way I do it is to get a common denominator on the RHS. Then coefficients for every power of $x$ must match — Andrei, Aug 10 '21 at 13:19
@jameselmore How do you get $(x-1)$ in the denominator if $A=0$? — Andrei, Aug 10 '21 at 13:20
It may simplify things to write $(x-1)(x^2+x+1)^2=(x^3-1)^2/(x-1)$ and get a simpler denominator $(x^3-1)^2$ — lhf, Aug 10 '21 at 13:26

score 4 · Answer 1 · answered Aug 10 '21 at 13:55

Not saying this is the easiest nor the most elegant one, but here's how I usually solve this kind of problem:

$$ \begin{align} \frac{x^{2}+3x+2}{\left(x-1\right)\left(x^{2}+x+1\right)^{2}}&=\frac{\left(x-1\right)\left(x+4\right)+2}{\left(x-1\right)\left(x^{2}+x+1\right)^{2}}\\ \\ &=\frac{x+4}{\left(x^{2}+x+1\right)^{2}}+\frac{2}{\left(x-1\right)\left(x^{2}+x+1\right)^{2}}\\ \\ &=\frac{x+4}{\left(x^{2}+x+1\right)^{2}}+\frac{\frac{2}{3}\left[\left(x^{2}+x+1\right)-\left(x-1\right)\left(x+2\right)\right]}{\left(x-1\right)\left(x^{2}+x+1\right)^{2}}\\ \\ &=\frac{\frac{1}{3}x+\frac{8}{3}}{\left(x^{2}+x+1\right)^{2}}+\frac{2}{3}\frac{1}{\left(x-1\right)\left(x^{2}+x+1\right)}\\ \\ &=\frac{\frac{1}{3}x+\frac{8}{3}}{\left(x^{2}+x+1\right)^{2}}+\frac{2}{3}\frac{\frac{1}{3}\left[\left(x^{2}+x+1\right)-\left(x-1\right)\left(x+2\right)\right]}{\left(x-1\right)\left(x^{2}+x+1\right)}\\ \\ &=\frac{\frac{1}{3}x+\frac{8}{3}}{\left(x^{2}+x+1\right)^{2}}+\frac{\frac{2}{9}}{x-1}+\frac{-\frac{2}{9}x-\frac{4}{9}}{x^{2}+x+1} \end{align} $$

That is a good one too, thanks! – King Crimson Aug 10 '21 at 13:57 — King Crimson, Aug 10 '21 at 13:57

score 3 · Accepted Answer · answered Aug 10 '21 at 13:50

3

Another easy coefficient is $M$. Multiply both sides by $x$ and take the limit as $x\to +\infty$: $$0=A+M\implies M=-A=-2/9$$ For $x=0$ we find $$2=-A+N+Q\implies N+Q=A+2=20/9.$$ For $x=i$ we find $$-3=\frac{A}{i-1} + \frac{Mi+N}{i} + \frac{Pi+Q}{-1}\implies Q=8/3, N+P=-1/9$$ Therefore $N=20/9-Q=-4/9$ and $P=-N-1/9=1/3$.

answered Aug 10 '21 at 13:50

Robert Z

145,942

Very elegant, thank you! – King Crimson Aug 10 '21 at 13:51

King Crimson · Answer 3 · 2021-08-10T14:17:27.493

I have been thinking about it for around 15 minutes and couldn't see any elegant or very short solutions. So decided to use pretty common and obvious method, that i already used before. I multiplied both sides of equation by $(x-1)(x^2+x+1)^2$ and here is what i got: $$x^2+3x-2=\frac{2}{9}(x^2+x+1)^2+(Mx+N)(x-1)(x^2+x+1)+(Px+Q)(x-1)$$ Then i got system of four equations(according to coefficients of the first polynomial) and solved them. Got: $M=-\frac{2}{9}, N=-\frac{4}{9}, P=\frac{1}{3}, Q=\frac{8}{3}$

But I didn't really wanted to use it, because i was told by multiple people, that it is not elegant and very straightforward

Partial fraction expansion question

3 Answers3