Find the integral $I=\int\frac{x^5-x^2-1}{x^5+x^4+x^3+x^2+x+1}dx$

Question

Find the integral $$I=\int\dfrac{x^5-x^2-1}{x^5+x^4+x^3+x^2+x+1}dx$$

Let $\dfrac{x^5-x^2-1}{x^5+x^4+x^3+x^2+x+1}=f(x).$ We can write $f(x)$ as follows $$f(x)=1+\dfrac{-x^4-x^3-2x^2-x-2}{x^5+x^4+x^3+x^2+x+1},$$ so the integral is actually $$I=\int dx+\int\dfrac{-x^4-x^3-2x^2-x-2}{x^5+x^4+x^3+x^2+x+1} dx=x+I_1$$

Now let $\dfrac{-x^4-x^3-2x^2-x-2}{x^5+x^4+x^3+x^2+x+1}=g(x)$. Then I tried to write $g(x)$ as $$g(x)=\dfrac{A}{x+1}+\dfrac{Bx+C}{x^2+x+1}+\dfrac{Dx+E}{x^2-x+1}\iff \\-x^4-x^3-2x^2-x-2=A(x^2+x+1)(x^2-x+1)+(Bx+C)(x+1)(x^2-x+1)+\\+(Dx+E)(x+1)(x^2+x+1)$$ Let us put $x=-1$ to get $$-1+1-2+1-2=A\cdot 3\Rightarrow A=-1$$

I don't see how to find the other unknowns $B,C,D,E$. My try:

Let $x_0$ be such that $x_0^2-x_0+1=0$. Then we want $$(*) -x_0^4-x_0^3-2x_0^2-x_0-2=(Dx_0+E)(x_0+1)(x_0^2+x_0+1)$$ Note that $$(x_0^2-x_0+1)^2=0=x_0^4-2x_0^3+3x_0^2+2x_0+1\Rightarrow -x_0^4+2x_0^3-3x_0^2+2x_0-1=0$$ Well then we can write the LHS of (*) as $$-x_0^4+2x_0^3-3x_0^2+2x_0-1-3x_0^3+x_0^2-3x_0-1\\=-3x_0^3+x_0^2-3x_0-1=-3x_0^3+x_0^2-x_0+1-2x_0-2=-3x_0^3-2x_0-2.$$ The RHS of (*) is $$(Dx_0+E)(x_0+1)(x_0^2-x_0+1+2x_0)\\=2x_0(x_0+1)(Dx_0+E)=2Dx_0^3+2(E+D)x_0^2+2Ex_0,$$ which cannot be true as there's no constant term here (for the LHS I got $-2$). I don't see where my mistake is. I tried to put $s:s^2+s+1=0$ and I reached something similar (that they cannot be equal). Any help would be appreciated.

Answer 1

Let me first mention the issue in your solution, then point out a way to simplify the problem to help preserve your sanity, and then I'll talk about the general method.

1. The issue with your solution is that you proposed using an $x_0$ that is a solution to $x^2-x+1$, which means $x_0$ is a number. Therefore, your concern:

which cannot be true as there's no constant term here (for the LHS I got −2)

is not actually a concern since $x_0$ IS a constant, so, in fact, there's nothing here that isn't a constant. Both sides are just numbers. (While we're on that subject, if we're working in $\mathbb{R}$ ... whoops! That polynomial has no real roots, so you can't do that. You may well point out that you could choose to work over $\mathbb{C}$, though, which is a valid point - and you could work out that solution and see that it's particularly awful because you've got a lot of $i$s running around.)

2. Let's so some simplifying. You figured out $A=-1$. A general policy with these problems is that, if you find something, ask yourself whether plugging it in will simplify your life. Since you're otherwise stuck, try it! $${−^4−^3−2^2−−2 \over(x+1)(^2++1)(^2−+1)} ={-1\over x+1} +{+ \over ^2++1}+{+ \over ^2-+1}$$ Hey, look at that, now you know a full term on the right side, so do a little work to move it to the left (again, this is general policy on these problems: keep your unknowns on the right and your knowns on the left). $${-x^3−x^2−x−1\over(x+1)(x^2+x+1)(x^2−x+1)} = {+ \over ^2++1}+{+ \over ^2-+1}$$ Hey, look at that, you've already done this factoring problem once today, so cancel that $x+1$: $${−x^2−1\over(x^2+x+1)(x^2−x+1)} = {+ \over ^2++1}+{+ \over ^2-+1}.$$

3. OK, now we're back with the difficulty you ran into before, which is that we still need to know how to deal with the remaining four unknowns; however, the problem is much simpler, so it'll be easier to do the algebra. In my opinion, what we've done so far definitely improved things (but maybe I'm biased because I make SO many errors when I do these calculations by hand that it's 100% worth it to me to simplify first - if I were using a computer/calculator to solve these, I wouldn't bother). Anyway, your method is a little too ad hoc for me; I prefer a more algorithmic method. First, I'll multiply through to get rid of the denominators, just like you did above: $$−x^2−1 = (Bx+C)(x^2-x+1)+(Dx+E)(x^2+x+1)$$ and now I'll take it one step farther and actually multiply out: $$−x^2−1 = Bx^3-Bx^2+Bx+Cx^2-Cx+C+Dx^3+Dx^2+Dx+Ex^2+Ex+E.$$ This looks like garbage. But, hey, maybe we could collect the terms on the right the way we normally write polynomials: $$−x^2−1 = (B+D)x^3+(-B+C+D+E)x^2+(B-C+D+E)x+(C+E).$$ Now a rule about polynomials - if two of them are the same, then their coefficients have to match perfectly. If you find that uncomfortable, don't worry, you're not alone; look at this one for intuition: if $3x^2 + 1 = 3x^2 + a$, then $a=1$, right? That's the only way you'd get the same values on the left and right all the time (ex, plug in $x=0$). Similarly, if $bx^2 +1 = 3x^2 + 1$, then $b=3$ (you could plug in $x=1$ and solve, or you could subtract 1 first, etc, etc). But back to your problem. Just notice that the constant terms on the left and right must match, and the coefficients of the first degree terms must also match, etc. We get a system of linear equations: $$B+D = 0\\ -B+C+D+E = -1\\ B-C+D+E = 0\\ C+E = -1$$ Solve these in your preferred fashion, and you're done! (Well, except for the completing the square or whatever else you need to do to take that particular integral.)

The solutions of the system are namely $$B=D=0,C=E=-\dfrac{1}{2}.$$

Answer 2

By expanding the polynomial expression that you have found you obtain

$$ (A+B)x^4+(C+E)x^3+(A+2(D+E))x^2+(B+2D+E)x+A+C+E=-x^4-x^3-2x^2-x-2 $$ i.e. $$ \begin{align} &A+B+D=-1\\ &C+2D+E=-1\\ &A+2D+2E=-2\\ &B+D+2E=-1\\ &A+C+E=-2 \end{align} $$ This is a system of 5 linear equation in 5 unknowns. Its solution is $$ A=-1,B=0,C=-\frac{1}{2},D=0,E=-\frac{1}{2} $$ The problem with your attemp is that, if you go back to your original equation you have $$ \frac{Dx+E}{x^2-x+1} $$ When you multiply this quantity for $x^2-x+1$ you are assuming that such polynomial is not zero. Moreover, when you evaluate a polynomial you obtain a number, so it doesn't really makes sense to talk about a "constant term" of $$ -3x_0^3-2x_0-2 $$ As it is a number, not a polynomial