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**symstar** $\displaystyle \int\frac{1}{x^3-1}dx$

Fraction work:

$\displaystyle \frac{1}{(x-1)(x^2+x+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+1}$

$\displaystyle 1=(A+B)x^2+(A-B+C)x+(A-C)$

$\displaystyle A=\tfrac{1}{3}; B=\tfrac{-1}{3}; C=\tfrac{2}{3}$

Back to the integral:

$\displaystyle \frac{1}{3}\int\frac{dx}{x-1}+\int\frac{\tfrac{-1}{3}x+\tfrac{2}{3}}{x^2+x+1}dx$

$\displaystyle \tfrac{1}{3}\ln{\left|x-1\right|}-\frac{1}{3}\int\frac{xdx}{x^2+x+1}+\frac{2}{3}\int \frac{dx}{x^2+x+1}$

I considered $\displaystyle u=x^2+x+1 \text{ ,but that gives } du=2x+1dx$ which doesn't seem to help the situation much. Should completing the square be a consideration?