Intuition for why the difference between $\frac{2x^2-x}{x^2-x+1}$ and $\frac{x-2}{x^2-x+1}$ is a constant? Why is the difference between these two functions a constant?
$$f(x)=\frac{2x^2-x}{x^2-x+1}$$
$$g(x)=\frac{x-2}{x^2-x+1}$$
Since the denominators are equal and the numerators differ in degree I would never have thought the difference of these functions would be a constant. 
Of course I can calculate it is true: the difference is $2$, but my intuition is still completely off here. So, who can provide some intuitive explanation of what is going on here? Perhaps using a graph of some kind that shows what's special in this particular case?
Thanks!

BACKGROUND: The background of this question is that I tried to find this integral: 
$$\int\frac{x dx}{(x^2-x+1)^2}$$
As a solution I found:
$$\frac{2}{3\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{2x^2-x}{3\left(x^2-x+1\right)}+C$$
Whereas my calculusbook gave as the solution:
$$\frac{2}{3\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{x-2}{3\left(x^2-x+1\right)}+C$$
I thought I made a mistake but as it turned out, their difference was constant, so both are valid solutions.
 A: Would you be surprised that the difference of $\dfrac{2x^2+x+1}{x^2}$ and $\dfrac{x+1}{x^2}$ is $2$?
A: It is just a bit of clever disguise.  Take any polynomial $p(x)$ with leading term $a_n x^n$.  
Now consider 
$$\frac{p(x)}{p(x)}$$
This is clearly the constant $1$ (except at zeroes of $p(x)$).  
Now separate the leading term:
$$\frac{a_n x^n}{p(x)} + \frac{p(x) - a_n x^n}{p(x)}$$
and re-write to create the difference:
$$\frac{a_n x^n}{p(x)} - \frac{a_n x^n - p(x)}{p(x)}$$
Obviously the same thing and hence obviously still $1$ but the first has a degree $n$ polynomial as its numerator and the second a degree $n - 1$ or less polynomial.  
Similarly, you could split $p(x)$ in many other ways.  
A: This belongs to a specific set of questions "you cannot really answer to your students" if you are a teacher.    
The difference is $2=\frac{2(x^2-x+1)}{x^2-x+1}$. This is why this seems weird (but true). Even this seems weird: $$\frac{15}{7}-\frac{1}{7}=2$$
The numerators differ by 14 (not 2) but the denominators are equal. 
The best is to try and explain this to yourself why this perfectly fine.
A: Note that in general given a rational function
$$f(x)=\frac{p(x)}{q(x)}\implies g(x)=f(x)+c=\frac{p(x)+c\cdot q(x)}{q(x)}$$
and $\deg(p(x)+c\cdot q(x))\le \max\{\deg(p(x)),\deg(q(x))\}$.
A: $$(2x^2-x)-(x-2)=2x^2-2x+2$$
Hence$$f(x)-g(x)=2\bigg(\frac{x^2-x+1}{x^2-x+1}\bigg)=2(1)=2$$
In effect, this pair of equations is a very specific case where the numerator and denominator end up lining up, and thus you get a constant for all values. 
A: 
Since the denominators are equal and the numerators differ in degree I would never have thought the difference of these functions would be a constant. 

When the numerators differ in degree, the difference between the numerators is a nontrivial polynomial that goes to infinity as $x\to\pm\infty$.
However, nobody says that this polynomial cannot grow at the same rate as the (equal) denominators, in which case the ratio will be bounded (and in particular possibly constant).  There's nothing in your observation "the denominators are equal and the numerators differ in degree" that link the denominator to either the numerators or their difference, so you have no reason to think the difference in numerators should dominate the growth of the entire fraction.
A: Maybe instead of integrating you need just differentiating.
$$\frac{df}{dx}=\frac{(2x^2-x)(2x-1)-(x^2-x+1)(4x-1)}{denom^2}\\
 =\frac{x^2-4x+1}{denom^2}$$
In the other hand:
$$\frac{dg}{dx}=\frac{(x-2)(2x-1)-(x^2-x+1)}{denom^2}\\
\ =\frac{x^2-4x+1}{denom^2}=\frac{df}{dx}$$
See that both functions have same derivates, which means they differ by the same value from $x$ to $x'$ , this only means they have a same growth rate and they have same difference all along the range of x axis.

A: I'm not sure anyone is speaking to your observation that the two numerators have different degrees.
Let's flip this around the other way:
\begin{align*}
\frac{x-2}{x^2-x+1} + 2 &= \frac{x-2}{x^2-x+1} + 2\frac{x^2-x+1}{x^2-x+1}  \\
  &= \frac{2x^2 -x}{x^2-x+1}  \text{.}
\end{align*}
That is, we started with a thing having a linear numerator and added a constant to it.  But when we brought the constant to have a common denominator, it picked up a degree two factor.  Then the addition was forced to produce a degree two sum.
To sum up, in the context of rational functions, when you add constants, you are adding polynomials having the degree of the denominator to the polynomials in the numerators.  So constants effectively have "degree two in the numerator" in your example.
A: You can construct a similar puzzle from the (simpler) observation that
$$
\frac{x}{x+1} + \frac{1}{x+1} = 1 .
$$
Translating, that's the same as 
$$
\frac{u-1}{u} + \frac{1}{u} = 1
$$
which isn't surprising at all.
A: Another way of stating what everyone has said:
$2x^2 - x= 2(x^2-x+1) + x - 2$
$\Rightarrow 2x^2-x = x-2 \mod x^2-x+1$
A: Because if $P\ and\ Q\ are\ polynomials$ then,
$$\deg \left( P\pm Q \right)\le max\left( \deg \left( P \right),\deg \left( Q \right) \right)$$
 where $deg$ is the degree of the polynomial. See this link
A: Here's another view on the problem:
Let's look at the fraction with the higher degree:
$$f(x)=\frac{2x^2-x}{x^2-x+1}$$
We note that the degree of the numerator is the same as the one of the denominator.
Now whenever the numerator's degree is at least as large as the denominator's degree, we can do polynomial division. The result is a polynomial whose degree is the difference between the numerator's and the denominator's degree, and a rest whose degree is less than the denominator's degree.
Now since numerator and denominator have the same degree, the difference of the degrees is zero. But a polynomial of degree zero is a constant. And the division rest will be of lesser degree.
Let's do the polynomial division explicitly:
$$
\begin{array}{llccl} &(2x^2-\ \ x) & / & (x^2-x+1) &=& 2\\ -&\underline{(2x^2-2x+2)}\\  & \ \ \ \ \ \ \ \ \ \ \ \ \ x \ - 2 \end{array}
%\begin{array}{rrrrcccl}
%&(2x^2 & -x) & & / & (x^2-x+1)  & = & 2\\
%-&(2x^2 & -2x & +2)\\
%\hline
% &      & x & -2
%\end{array}
$$
You surely will recognize the rest as the numerator of $g(x)$
So, whenever you have a rational function whose numerator and denominator, after cancelling out, have the same non-zero degree, it is not only possible, but guaranteed that there exists another rational function with lesser degree in the numerator which differs from the original rational function only by a constant.
A: I feel that it's only natural that intuition would fail you in cases like these. If you wrote the two functions out as
$$f(x)=\frac{2\left(x^2-x+1\right)+\left(x-2\right)}{x^2-x+1}$$
$$g(x)=\frac{x-2}{x^2-x+1}$$
it would be obvious that the difference between them was 2. This is the first thing you could trip over, the degree of the numerator don't actually differ by 1, they are the same.
I guess if you made it a rule to divide whenever you have a rational function where the degree of the numerator is greater than or equal to the degree of the denominator until it isn't, you would get consistent results. (Kinda like how you learn waaay back how $\frac{13}{7}$ is an improper fraction, because it's actually bigger than a whole number, and that you should write it as $1\frac{6}{7}$. a function $f(x)=\frac{P(x)}{Q(x)}$ where the degree of $P(x)$ is greater than or equal to the degree of $Q(x)$ is called an improper rational function.)
But, and this might be another thing that's bugging you about this question, you can't just do that! (As my algebra teacher used to exclaim in exasperation.) You introduce a new constraint: $x^2-x+1\neq0$. Which isn't relevant if x is real, but does change things up a bit if  it s a complex number.
You have asked for graphs, and ye shall receive.
This is probably what you get when you graph the two functions $f$ and $g$.

But if you allow for complex values of $x$, it can also look like this:

(Sans the vertical line at x=0.5, I don't know why that is happening, probably a limitation in the graphing engine on desmos.)
Not so obvious that the difference between the two is 2 in this case...
Sorry, I just realized this probably doesn't answer your question on how you can reconciliate this apparent discrepancy with your intuition, but I thought explaining why it may not be that clear-cut would help at least a little bit.
Playing with the parameters used in the graph might help further your understanding, here's a link to the interactive graph, with a little more in depth graphical interpretation of why you might doubt what you logically "know" to be true.
As for the equations I used in place of the given ones... You'll just have to trust my algebra (which you shouldn't in critical situations, but I think this is safe enough hehe). I did them on a legal pad, but I'll type them up if anyone insists.
A: Look at it in reverse:
Take a polynomial fraction and add it to a constant.  The result will be a polynomial fraction, with the same denominator and a different polynomial as the numerator.
A: Since the denominator is the same , you subtract one numerator from the other. In this special case you find the resultant numerator is twice the denominator. Feels like that was much simpler than the rest of your background calculus.
A: I shall try to explain an intuitive way to think about this.
Disclaimer: I am only developing on @gandalf61's answer.
It must be very clear to you that $11/3$ and $2/3$ defer by an integer. Why is this so? Because in the world $``$modulo 3$"$, 11 and 2 are but the same. In the same vein, we can say that in the world of $``$modulo $x^2-x+1"$, $2x^2-x$ and $x-2$ are but the same because $2x^2-x=2(x^2-x+1)+x-2$. When we are in such a world, the same polynomial may appear in different forms with different "degrees". So in a world such as the ideal generated by $x^2-x+1$ (ie $<x^2-x+1>$ ), degree is no longer retained. One's intuition may fail here because one is not familiar with such a world where degree is not so well behaved.
I hope this addresses your question. Thank you.
