The given problem is $\int{x\over x^3-1}dx$.
I know this equals
${1\over3}\int {1\over x-1}-{x-1\over x^2+x+1}dx,$ which can be separated into ${1\over3}\int {1\over x-1}dx - {1\over3}\int{x+(1/2)-(3/2)\over x^2+x+1}dx.$ This can further be separated into ${1\over3}\int{1\over x-1}dx - {1\over3}\int{x+(1/2)\over x^2+x+1}dx + {1\over2}\int{1\over(x+(1/2))^2+(3/4)}dx.$
I know the integral of ${1\over3}\int {1\over x-1}dx$ is $(1/3)\ln(x+1)+C$ where $C$ is an arbitrary constant. Using u-subsitution, where $u=x^2+x+1$ and $du=(2x+1)dx$ and $(1/2)du=(x+(1/2))dx$, I know the integral of $-{1\over3}\int{ x+(1/2)\over x^2+x+1 }dx$ is $(1/6)\ln(x^2+x+1)+C$. I need to get the last part, ${1\over2}\int{1\over(x+(1/2))^2+(3/4)}dx$ to be some form of arctan. I can use u-substitution where $u=x+(1/2)$ and $du=dx$, but I don't know where to go from there.