TITLE: Integration with $\ln(x)$ in the denominator
QUESTION [13 upvotes]: Find$$\displaystyle\int_1^\infty\frac{(x^2-1)(x^4-1)(x^6-1)}{\ln(x)(x^{14}-1)} dx$$

I tried simplifying the terms without logarithm 
$x^2-1=(x-1)(x+1)\\x^{14}-1=(x^7-1)(x^7+1)$
to see if any substitution is possible but since the logarithm is in the denominator it is getting hard and I am not sure if there is any substitution that would work.
I am looking for hints to get started

REPLY [7 votes]: Via the substitution $x=e^{y}$ we get that
\begin{equation*}
I= \int_{1}^{\infty}\dfrac{\left(x^2-1\right)\left(x^4-1\right)\left(x^6-1\right)}{\ln(x)(x^{14}-1)}\, dx = \int_{0}^{\infty}\dfrac{\left(e^{-2y}-1\right)\left(e^{-4y}-1\right)\left(e^{-6y}-1\right)e^{-y}}{y\left(e^{-14y}-1\right)}\, dy.
\end{equation*}
If
\begin{equation*}
 f(y) = \dfrac{e^{-y}+e^{-2y}-e^{-3y}+e^{-4y}-e^{-5y}-e^{-6y}}{e^{-7y}-1}
\end{equation*}
then
\begin{equation*}
 f(2y)-f(y) =\dfrac{\left(e^{-2y}-1\right)\left(e^{-4y}-1\right)\left(e^{-6y}-1\right)e^{-y}}{e^{-14y}-1}
\end{equation*}
and
\begin{equation*}
 I = \int_{0}^{\infty}\dfrac{f(2y)-f(y)}{y}\, dy.
\end{equation*}
Since $\lim_{y\to 0}f(y)=-1$ and $\lim_{y\to +\infty}f(y)=0$ we have a Frullni integral. 
See https://en.wikipedia.org/wiki/Frullani_integral
Consequently 
\begin{equation*}
 I=-\ln\dfrac{1}{2}=\ln 2.
\end{equation*}