<s>[INST] 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.
 [/INST] 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.

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