Answer
$\frac{1}{3}\ln|x-1|-\frac{1}{6}\ln|x^2+x+1|-\frac{1}{\sqrt 3}\tan^{-1}\left(\frac{2x+1}{\sqrt 3}\right)+C$
Work Step by Step
$\int\frac{1}{x^3-1}dx$ = $\int\frac{1}{(x-1)(x^2+x+1)}dx$ = $\int\left(\frac{A}{x-1}+\frac{Bx+C}{x^2+x+1}\right)dx$
$1$ = $A(x^2+x+1)+(Bx+C)(x-1)$
$1$ = $A(x^2+x+1)+(Bx^2-Bx+Cx-C)$
$0$ = $A+B$
$0$ = $A-B+C$
$1$ = $A-C$
$A$ = $\frac{1}{3}$, $B$ = $-\frac{1}{3}$, $C$ = $-\frac{2}{3}$
So
$\int\frac{1}{x^3-1}dx$ = $\int{\frac{1}{3}}\left[\frac{1}{x-1}-\frac{x+2}{x^2+x+1}\right]dx$ = $\int{\frac{1}{3}}\left[\frac{1}{x-1}-\frac{x+{\frac{1}{2}}}{x^2+x+1}-\frac{{\frac{3}{2}}}{(x+\frac{1}{2})^2+{\frac{3}{4}}}\right]dx$ = $\frac{1}{3}\ln|x-1|-\frac{1}{6}\ln|x^2+x+1|-\frac{1}{\sqrt 3}\tan^{-1}\left(\frac{x+{\frac{1}{2}}}{\frac{\sqrt 3}{2}}\right)+C$ = $\frac{1}{3}\ln|x-1|-\frac{1}{6}\ln|x^2+x+1|-\frac{1}{\sqrt 3}\tan^{-1}\left(\frac{2x+1}{\sqrt 3}\right)+C$
