<s>[INST] Finding $\int\frac{2x^3-1}{x^6-x^4+2x^3-2x^2-x+1}dx$ Finding $\displaystyle \int\frac{2x^3-1}{x^6-x^4+2x^3-2x^2-x+1}dx$
Attempt: Assume $\displaystyle \int\frac{2x^3-1}{x^6-x^4+2x^3-2x^2-x+1}dx =\int \frac{2x-x^{-2}}{\bigg[x^4-x^2+2x-\frac{2}{x}-\frac{1}{x}+\frac{1}{x^2}\bigg]}dx$
could some help me how to  solve,thanks
 [/INST] $\dfrac{2x - \frac{1}{x^2}}{x^4 -x^2 +2x -2 -\frac{1}{x} + \frac{1}{x^2}}$
We want to write the denominator in $x^2 + \frac{1}{x}$ since it is derivative in the nominator
Reordering
$$x^4 +2x +\frac{1}{x^2}-2  -x^2 - \frac{1}{x} = \left(x^2 +\frac{1}{x}\right)^2 - 2 -(x^2 + \frac{1}{x}) $$
Let $ u = x^2 + \frac{1}{x^2}$ 
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