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

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 $\ln 2$.
