I tried adding and subtracting 1 to bring a square expression with numerator as $(x^3-1)^2 -1$ but always going to partial fraction which obviously is very lengthy considering the cubic factor in denominator.


Let $\displaystyle I = \int\frac{x^6-2x^3}{(x^3+1)^3}dx = \int\frac{x^6-2x^3}{x^6\left(x+\frac{1}{x^2}\right)^3}dx = \int\frac{1-2x^{-3}}{\left(x+\frac{1}{x^2}\right)^3}dx$

Now Put $\displaystyle x+\frac{1}{x^2} = t\;,$ Then $\displaystyle \left(1-\frac{2}{x^3}\right)dx = dt$

So Integral $\displaystyle I = \int\frac{1}{t^3}dt = -\frac{1}{2t^2}+\mathcal{C} = -\frac{1}{2}\cdot \left(\frac{x^2}{x^3+1}\right)^2+\mathcal{C}$

  • $\begingroup$ The denominator of the integrand should be $(x^3+1)^3$. $\endgroup$ – Empiricist Sep 13 '15 at 10:26
  • $\begingroup$ thanks SRX I have edited it. $\endgroup$ – juantheron Sep 13 '15 at 10:28
  • $\begingroup$ Dang it i always go after the wrong substitutions $\endgroup$ – Sujith Sizon Sep 13 '15 at 10:42
  • 2
    $\begingroup$ could not be shorter! $\endgroup$ – Math-fun Nov 26 '15 at 10:10

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