Fastest way to integrate $\frac{1}{t(t^6-1)}$ My friend mentioned to me that there is a very quick way to integrate $\frac{1}{t(t^6-1)}$. 
The standard method for dealing with integrals of rational functions by partial fraction decomposition is long (but doable) as there are 5 terms to integrate. After playing around for some time I still don't see the fast way of doing this integral.
Am I missing something obvious?
 A: Note that
$$\frac1{t(t^6-1)}=\frac{t^6-(t^6-1)}{t(t^6-1)}=\frac{t^6}{t(t^6-1)}-\frac{t^6-1}{t(t^6-1)}=\frac{t^5}{t^6-1}-\frac1t$$
The first fraction is just a u-substitution and the second is the natural logarithm, hence
$$\int\frac1{t(t^6-1)}\ dt=\frac16\ln|t^6-1|-\ln|t|+c$$

For a more general method, we have PFD over the complex plane:
$$\frac1{t(t^6-1)}=\sum_\omega\frac{\operatorname{Res}(f,\omega)}{t-\omega}$$
where the sum is taken over the roots of $t(t^6-1)$.  When $\omega=0$, we have
$$\operatorname{Res}(f,0)=\lim_{t\to0}\frac t{t(t^6-1)}=-1$$
When $\omega^6=1$, we have
$$\operatorname{Res}(f,\omega)=\lim_{t\to\omega}\frac{t-\omega}{t(t^6-1)}\stackrel{\text{L'H}}=\frac16$$
Thus, it becomes clear that
$$\int\frac1{t(t^6-1)}\ dt=-\ln|t|+c+\frac16\sum_{\omega^6=1}\ln|t-\omega|=\frac16\ln|t^6-1|-\ln|t|+c$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\int{\dd t \over t\pars{t^{6} - 1}} &
\,\,\,\stackrel{t\ =\ x^{-2}}{=}\,\,\,
2\int{x^{11} \over x^{12} - 1}\,\dd x = {1 \over 6}\,\ln\pars{x^{12} - 1} =
{1 \over 6}\,\ln\pars{{1 \over t^{6}} - 1}
\\[5mm] & =
\bbx{\ds{{1 \over 6}\,\ln\pars{1 - t^{6}} - \ln\pars{t} +
\pars{~\mbox{a constant}~}}}
\end{align}
