Answer
$ \displaystyle \frac{1}{3}\ln|x-1| -\frac{1}{6}\cdot\ln|x^{2}+x+1|-\frac{1}{\sqrt{3}}\arctan(\frac{2x+1}{\sqrt{3}})+C$
Work Step by Step
$x^{3}-1$ is a difference of cubes
$\displaystyle \frac{1}{(x-1)(x^{2}+x+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^{2}+x+1}$
$1=A(x^{2}+x+1)+(Bx+C)(x-1)$
$1=Ax^{2}+Ax+A+Bx^{2}-Bx+Cx-C$
$1=(A+B)x^{2}+(A-B+C)x+(A-C)$
$A+B=0\Rightarrow A=-B$
$A-C=1\Rightarrow C=A-1$
$A-B+C=0\displaystyle \Rightarrow A+A+A-1=0\Rightarrow A=\frac{1}{3}$
$B=-\displaystyle \frac{1}{3},C=-\frac{2}{3}$
$\displaystyle \frac{1}{(x-1)(x^{2}+x+1)}=\frac{\frac{1}{3}}{x-1}+\frac{-\frac{1}{3}x-\frac{2}{3}}{x^{2}+x+1}$
$=\displaystyle \frac{1}{3(x-1)} -\frac{x+2}{3(x^{2}+x+1)}$
$=\displaystyle \frac{1}{3(x-1)} -\frac{1}{3}\cdot\frac{\frac{1}{2}(2x+1)+\frac{3}{2}}{ x^{2}+x+1}$
$=\displaystyle \frac{1}{3(x-1)} -\frac{1}{6}\cdot\frac{2x+1}{ x^{2}+x+1} -\frac{1}{3}\cdot\frac{\frac{3}{2}}{ x^{2}+x+(\frac{1}{2})^{2}+\frac{3}{4}}$
$=\displaystyle \frac{1}{3(x-1)} -\frac{1}{6}\cdot\frac{2x+1}{ x^{2}+x+1} -\frac{1}{2}\cdot\frac{1}{ (x+\frac{1}{2})^{2}+(\frac{\sqrt{3}}{2})^{2}}$
$\displaystyle \int\frac{1}{x^{3}-1}dx=$
$\displaystyle \frac{1}{3}\ln|x-1| -\frac{1}{6}\cdot\ln|x^{2}+x+1|-\frac{1}{\sqrt{3}}\arctan(\frac{2x+1}{\sqrt{3}})+C$
