The polynomial $x^2 - x + 1$ has discriminant $(-1)^2 - 4(1)(1) = -3 < 0$, so it has no real roots. According to the partial fractions recipe, there are unique real constants $A,B,C,D,E$ such that
$\frac{2x^2+1}{x^5+x^2} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1} + \frac{Dx+E}{x^2-x+1}$
for all $x \in \mathbb{R}$. In other words, you did not postulate a sufficiently general expression for the partial fractions decomposition, so -- except in the unlikely event that $A = D = 0$ -- you will not be able to achieve a decomposition in the form you have given.
I happened to cover partial fractions quite recently in a class I am teaching this semester. A student asked me why you need a term like $\frac{A}{x}$. My answer at the time was to think of taking a sum which did have a $\frac{A}{x}$ term, put it over a common denominator, and then apply a partial fractions decomposition: we would then be saying that there are some real constants $\alpha,\beta,\gamma,\delta$ such that
$\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1} + \frac{Dx+E}{x^2-x+1} = \frac{\alpha}{x^2} + \frac{\beta}{x+1} + \frac{\gamma x + \delta}{x^2-x+1},$
which seems unlikely. Actually I think I could say it a bit better: a general proper rational function with denominator $x^5+x^2$ is of the form
$\frac{ax^4 + bx^3 + cx^2 + dx + e}{x^5+x^2},$
thus there is in a sense a five parameter family of such things. Therefore whatever your partial fractions decomposition should be, it should also have five parameters $A,B,C,D,E$. That's a more convincing answer as to why your proposed expansion is (very probably) not achievable -- isn't it? -- you are trying to express a five parameter family of functions using only three parameters: no good.
This business with "parameters" can be made rigorous using the language of linear algebra, and in fact dimension-counting can be used to give a proof of the existence and uniqueness of the partial fraction decomposition. See this note which I wrote just a few days ago.