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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

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  • $\begingroup$ I would suggest factorising...Partial Fractions and then integrating $\endgroup$
    – LM2357
    Commented Mar 20, 2017 at 5:10
  • $\begingroup$ Hint: the denominator factors as $(x-1)(x^2+x-1)(x^3+x+1)$ $\endgroup$
    – Grant B.
    Commented Mar 20, 2017 at 5:11

3 Answers 3

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$\dfrac{2x - \frac{1}{x^2}}{x^4 -x^2 +2x -2 -\frac{1}{x} + \frac{1}{x^2}}$

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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  • $\begingroup$ Beautiful solution! +1 $\endgroup$ Commented Mar 20, 2017 at 13:13
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$$\int\frac{2x^3-1}{x^6-x^4+2x^3-2x^2-x+1}dx=\frac{1}{3}\int\left(\frac{1}{x-1}+\frac{2x+1}{x^2+x-1}-\frac{3x^2+1}{x^3+x+1}\right)dx=$$ $$=\frac{1}{3}\left(\ln|x-1|+\ln|x^2+x-1|-\ln|x^3+x+1|\right)+C$$

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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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