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.
 A: The purpose is to get an integral of the form $\displaystyle\int \frac{1}{v^{2}+1}\,dv=\arctan v$. Using the substitution indicated by you $u=x+1/2$, $du=dx$,
we obtain
$$\begin{eqnarray*}
\frac{1}{2}\int \frac{1}{\left( x+1/2\right) ^{2}+3/4}dx &=&\frac{1}{2}\int 
\frac{1}{u^{2}+3/4}\,du \\
&=&\frac{1}{2\cdot 3/4}\int \frac{1}{\frac{u^{2}}{3/4}+1}\,du \\
&=&\frac{2}{3}\int \frac{1}{\left( \frac{u}{\sqrt{3}/2}\right) ^{2}+1}\,du,
\end{eqnarray*}$$
where we have manipulated the integrand so that it is of the form $
1/(v^{2}+1)$, with $v=\dfrac{u}{\sqrt{3}/2}$. Using this new substitution,
since $dv=\frac{du}{\sqrt{3}/2}$, the integral becomes
$$\begin{eqnarray*}
\frac{2}{3}\int \frac{1}{\left( \frac{u}{\sqrt{3}/2}\right) ^{2}+1}\,du &=&%
\frac{2}{3}\int \frac{1}{v^{2}+1}\cdot \frac{\sqrt{3}}{2}\,dv \\
&=&\frac{\sqrt{3}}{3}\int \frac{1}{v^{2}+1}\,dv \\
&=&\frac{\sqrt{3}}{3}\arctan v \\
&=&\frac{\sqrt{3}}{3}\arctan \left( \frac{u}{\sqrt{3}/2}\right)  \\
&=&\frac{\sqrt{3}}{3}\arctan \left( \frac{x+1/2}{\sqrt{3}/2}\right)  \\
&=&\frac{\sqrt{3}}{3}\arctan \left( \frac{\sqrt{3}}{3}\left( 2x+1\right)
\right) +\text{Const.}
\end{eqnarray*}$$
Added. We could have used the single substitution $v=\dfrac{x+1/2}{\sqrt{3}/2}$,
resulting from both substitutions $v$ and $u$.
$$\begin{eqnarray*}
\frac{1}{2}\int \frac{1}{\left( x+1/2\right) ^{2}+3/4}dx &=&\frac{1}{2}\int 
\frac{2}{3}\frac{\sqrt{3}}{v^{2}+1}\,dv \\
&=&\frac{1}{3}\sqrt{3}\arctan v=\frac{1}{3}\sqrt{3}\arctan \left( \frac{x+1/2
}{\sqrt{3}/2}\right) +\text{ Const.}
\end{eqnarray*}$$
A: 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 $a\sqrt{1-w^2}$. 
