November
10, 2006
Evaluate
$\displaystyle\int \frac1{x^7-x}\,dx$.
Solution. We have
$\begin{eqnarray}
\int\frac1{x^7-x}\,dx &=& \int \frac{1}{x(x^6-1)}\,dx \\
&=& \int\frac{x^5}{x^6(x^6-1)}\,dx
\end{eqnarray}$
Now let $u=x^6$. Then
$\begin{eqnarray}
\int\frac1{x^7-x}\,dx &=& \frac16\int\frac{1}{u(u-1)}\,du \\
&=& \frac16\int \left(\frac1{u-1} - \frac1{u} \right)\,du \\
&=& \textstyle\frac16\Big( \ln |u-1| - \ln |u| \Big) + C \\
&=& \textstyle\frac16 \ln \left| \displaystyle\frac{u-1}{u}
\right| + C \\
&=& \textstyle\frac16 \ln \left| \displaystyle\frac{x^6-1}{x^6}
\right| + C.
\end{eqnarray}$