Integration with $\ln(x)$ in the denominator 
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
 A: 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*}
A: The following answer is similar to Song's answer except that it involves using a common integral representation of the digamma function:
$$\psi(z) = -\gamma + \int_{0}^{1} \frac{1-t^{z}}{1-t} \, \mathrm dt\, , \quad \mathcal \Re (z) >0$$
The integral representation of the tetragamma function used in Song's answer can be derived from this integral representation by differentiating under the integral sign twice.

$$ \begin{align} &\int_{1}^{\infty} \frac{(x^{2}-1)(x^{4}-1)(x^{6}-1)}{(x^{14}-1)\ln x} \, \mathrm dx \\ &= \int_{0}^{1}\frac{(1-u^{2})(1-u^{4})(1-u^{6})}{(1-u^{14}) \ln \frac{1}{u}} \, \mathrm du\\ & = \int_{0}^{1} \int_{0}^{6} \frac{(1-u^{2})(1-u^{4}) \, u^{a}}{(1-u^{14}) \ln u} \, \mathrm da \, \mathrm du\\ &= \int_{0}^{6} \int_{0}^{1} \frac{(1-u^{2})(1-u^{4}) \, u^{a}}{(1-u^{14}) } \, \mathrm du \,\mathrm da \\\ &= \frac{1}{14}\int_{0}^{6} \int_{0}^{1} \frac{(1-w^{2/14})(1-w^{4/14}) \, w^{a/14}}{1-w } \, w^{-13/14} \, \mathrm dw \, \mathrm da \\ &= \frac{1}{14}\int_{0}^{6} \int_{0}^{1} \left(\frac{w^{(a+1)/14-1}-w^{(a+5)/14-1}}{1-w} + \frac{w^{(a+7)/14-1} -w^{(a+3)/14-1}}{1-w} \right) \, \mathrm dw  \,\mathrm da \\&= \frac{1}{14} \int_{0}^{6}  \left[ \left(\psi \left(\frac{a+5}{14} \right) - \psi \left(\frac{a+1}{14} \right) \right) + \left(\psi \left(\frac{a+3}{14} \right) - \psi \left(\frac{a+7}{14} \right) \right) \right] \, \mathrm d a  \\ &= \ln \left[\frac{\Gamma \left(\frac{11}{14} \right) \Gamma\left(\frac{9}{14} \right)}{\Gamma \left(\frac{1}{2} \right)\Gamma \left(\frac{13}{14} \right)} \right]\,  - \ln \left[\frac{\Gamma \left(\frac{5}{14} \right)\Gamma \left(\frac{3}{14} \right)}{\Gamma \left(\frac{1}{14} \right)\Gamma \left(\frac{1}{2} \right)}\right] \\&= \ln \left[\frac{\Gamma \left(\frac{1}{14} \right)\Gamma \left(\frac{9}{14} \right) \Gamma\left(\frac{11}{14} \right)}{\Gamma \left(\frac{3}{14} \right)\Gamma \left(\frac{5}{14} \right)\Gamma \left(\frac{13}{14} \right)} \right] \end{align}  $$
See Song's answer for how to use properties of the gamma function to show that the above result reduces to $\ln 2$.
A: This is more of a long comment than an answer, but I think it's worth adding since it gets overlooked pretty often.
When possible, it's usually a good idea to use a computer to get a decent first approximation on calculation-heavy problems like this.
When I tried this problem, I started by inputting the integral into GeoGebra, then copying the number into Wolfram Alpha to search for a possible closed form. W.A. gave me $\ln(2)$ (which turned out to be correct), and from there I tried something similar to what Song did to verify that this was correct (except that I messed it up and ended up with a different answer).
The point is, sometimes it's easier to have a machine compute the right answer and then figure out how you would get that answer analytically, instead of trying to solve the problem by hand from the start.
Of course, you should never take the answer for granted - machines can always malfunction - but when it comes to raw calculations, computers tend to be faster and make fewer mistakes than humans. That's what calculators are there for, after all.
