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I am having quite a difficult time integrating

$$ \int \frac{\mathrm{d}x}{x^3-1} $$

My first approach was to apply a partial fraction decomposition

$$ \int \frac{\mathrm{d}x}{x^3-1} = \int \frac{\mathrm{d}x}{(x-1)(x^2+x+1)} = \frac{1}{3} \int \frac{\mathrm{d}x}{x-1} - \frac{1}{3} \int \frac{x+2}{x^2+x+1}\mathrm{d}x$$

The first term is simple enough to integrate but I can't figure out the second. I've tried each of the integration tricks in my arsenal, integration of parts gives a more complex equation, a trigonometric substitution doesn't seem to get me anywhere, and lastly I'm not sure what to do with a u-substitution since if $ u = x^2+x+1 $ then $ \mathrm{d}u = (2x+1)\mathrm{d}x \neq (x+2)\mathrm{d}x $. And further breaking up the formula further doesn't seem to eliminate these problems.

My last effort was to get a solution from Wolfram and try to reverse engineer it by taking the derivative

$$ \frac{\mathrm{d}}{\mathrm{d}x} \left( - \frac{1}{6} \ln (x^2+x+1) + \frac{1}{3} \ln (1-x) - \frac{\sqrt{3}}{3} \tan ^{ - 1} \left( \frac{2x+1}{\sqrt{3}} \right) \right) $$

from which I got

$$ \frac{1}{3(1-x)} - \frac{1}{2(x^2 + x + 1)} - \frac{2x+1}{6x^2 + 6x + 6} $$

However I can't seem to figure out the method or motivation for the fraction decomposition. I'm I missing something really obvious or am I missing a tool of integration? Thank you very much for your help!

From the first answer suggesting completing the square I believe I was able to figure the rest out. One then gets a difference of squares and can then use a trigonometric substitution where

$$ x + \frac{1}{2} = \frac{\sqrt{3}}{2} \tan \theta $$

the answer I got is

$$ \frac{1}{3} \ln |x-1| - \frac{1}{6} \ln(x^2+x+1) - \frac{\sqrt{3}}{3} \tan ^{ - 1} \left( \frac{2x+1}{\sqrt{3}} \right) + C $$

My last confusion is the discrepancy in the solutions. Wolfram's second term seems to be undefined over $ (1, \infty) $, whereas I believe that $$ \int \frac{\mathrm{d}x}{x^3-1} $$ should be defined over that interval.

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You can write $\displaystyle\int\frac{x+2}{x^2+x+1} dx=\frac{1}{2}\int\frac{2x+1}{x^2+x+1} dx + \frac{3}{2}\int\frac{1}{x^2+x+1} dx$

$\displaystyle=\frac{1}{2}\ln(x^2+x+1)+\frac{3}{2}\int\frac{1}{(x+\frac{1}{2})^2+\frac{3}{4}} dx$, and now let $u=x+\frac{1}{2}$.

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  • $\begingroup$ Thank you, I completely missed splitting the numerator $ x + 2 $ into $ x + \frac{1}{2} $ and $ \frac{3}{2} $. $\endgroup$ – user38770 Sep 14 '14 at 8:35
  • $\begingroup$ You're welcome. As indicated in the other answer, you could also complete the square first instead. If you do that, then letting $u=x+\frac{1}{2}$, $x=u-\frac{1}{2}$, $dx=du$ gives $\int\frac{x+2}{x^2+x+1} dx=\int\frac{u+\frac{3}{2}}{u^2+\frac{3}{4}} du$. $\endgroup$ – user84413 Sep 14 '14 at 20:01
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Hint: Complete the square in the denominator: $x^2+x+1 = (x + 1/2)^2 + 3/4$. Now write the fraction as a linear combination of two fractions whose integral you can do.

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