<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] As Grant B. commented
$$x^6-x^4+2x^3-2x^2-x+1=(x-1)(x^2+x-1)(x^3+x+1)$$ Then, Using partial fraction decomposition, 
$$\frac{2x^3-1}{x^6-x^4+2x^3-2x^2-x+1}=\frac 13 \left(\frac{2 x+1}{ x^2+x-1}-\frac{3 x^2+1}{
   x^3+x+1}+\frac{1}{ x-1}\right)$$ where you can notice that each numerator is  the derivative of each denominator.
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