Integrate using Partial Fraction decomposition, completing the square 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.

Use the substitution 
$$\frac{\sqrt{3}}{2}u=x+\frac{1}{2}.$$
Or else, if you want to do it in two steps, make the substitution $u=x+\frac{1}{2}$.
You will end up with an expression that includes $u^2+\frac{3}{4}$. Then let $u=\frac{\sqrt{3}}{2}v$.  The substitution that I proposed is a little faster.
Remark: Suppose that we want to find the integral 
$$\int\frac{dx}{x^2+k},$$
where $k$ is a positive constant. Write $k=a^2$, where $a$ is positive. Then $a=\sqrt{k}$. We are interested in
$$\int\frac{dx}{x^2+a^2}.$$
Let $x=aw$ (or equivalently, $w=\frac{x}{a}$). Then $dx=a\,dw$. Substituting, we get
$$\int \frac{a\,dw}{a^2w^2+a^2}, \quad\text{which is}\quad \int\frac{1}{a}\frac{dw}{w^2+1}.$$
More or less the same idea works, for example, when we want to integrate something that involves $\sqrt{k-x^2}$, where $k$ is positive. Write $k=a^2$, where we choose $a$ positive.  So we are interested in $\sqrt{a^2-x^2}$. Make the substitution $u=a\sin\theta$. Or else, make the preliminary substitution $x=aw$. Then our expression becomes $\sqrt{a^2-a^2w^2}$, which is equal to $a\sqrt{1-w^2}$. 
