Integration with $\ln(x)$ in the denominator

By making substitution $\frac 1 x\mapsto x$, we get \begin{align*} I = \int_0^1 \frac{(x^2-1)(x^4-1)(x^6-1)}{\ln x \ (1-x^{14})} dx. \end{align*} We can observe that $$ \int_0^\alpha x^r dr = \left[\frac{x^r}{\ln x}\right]^\alpha_0 = \frac{x^\alpha -1 }{\ln x} $$ hence \begin{align*} I =& \int_0^1 \left(\int_0^2 x^r dr\right) \left(\int_0^4 x^s ds\right) \left(\int_0^6 x^t dt\right)\frac{\ln^2 x}{1-x^{14}} dx\\ =& \int_0^6\int_0^4\int_0^2 \left( \int_0^1 \frac{ x^{r+s+t}\ln^2 x}{1-x^{14}} dx\right) \ drdsdt. \end{align*} For the inner integral, make substitution $x^{14}\mapsto x$ to get $$ \int_0^1 \frac{ x^{r+s+t}\ln^2 x}{1-x^{14}} dx = \frac 1 {14^3} \int_0^1 \frac{x^{\frac{r+s+t+1}{14}-1}\ln^2 x}{1-x} dx. $$ Now we can use the integral representation of polygamma function: $$ \psi''(s) = - \int_0^1 \frac{x^{s-1} \ln^2 x}{1-x} dx $$ where $\psi(s)=\frac{d}{ds}\ln \Gamma(s)$ is the digamma function. If we write $D^t f(x) :=f(x+t)-f(x)$, then we can write \begin{align*} I = & -\frac 1{14^3}\int_0^6\int_0^4\int_0^2 \psi''\left(\frac {r+s+t+1}{14} \right) drdsdt\\ =&-\frac 1{14^2}\int_0^6\int_0^4 \psi'\left(\frac {s+t+3}{14} \right)-\psi'\left(\frac {s+t+1}{14} \right) dsdt\\ =&-\frac 1{14^2}\int_0^6\int_0^4 D^{1/7}\psi'\left(\frac {s+t+1}{14} \right) dsdt\\ =&-\frac 1{14}\int_0^6D^{2/7} D^{1/7}\psi\left(\frac {t+1}{14} \right) dt\\ =& -\left[D^{2/7} D^{1/7}\ln \Gamma\left(\frac {t+1}{14} \right)\right]^{6}_{t=0}\\ =&-D^{3/7}D^{2/7}D^{1/7}\ln \Gamma\left(\frac 1 {14}\right) \\ =& \cdots \\ =&\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*}

Finally, using the duplication formula of Gamma function (thanks to @Metamorphy) $$ \Gamma(z)= 2^{1-2z}\pi^{\frac 1 2}\frac{\Gamma(2z)}{\Gamma\left(z+\frac 1 2 \right) } $$ and the fact that $\Gamma(z+1) = z\Gamma(z)$, we get:

For the numerator, \begin{align*} \Gamma\left(\frac {1}{14}\right)\Gamma\left(\frac {9}{14}\right)\Gamma\left(\frac {11}{14}\right)=& \pi^{\frac 32}\frac{\Gamma\left(\frac {2}{14}\right)\Gamma\left(\frac {18}{14}\right)\Gamma\left(\frac {22}{14}\right)}{\Gamma\left(\frac {8}{14}\right)\Gamma\left(\frac {16}{14}\right)\Gamma\left(\frac {18}{14}\right)}\\ =&\pi^{\frac 32}\frac{\Gamma\left(\frac {2}{14}\right)\Gamma\left(\frac {18}{14}\right)\frac 8 {14}\Gamma\left(\frac {8}{14}\right)}{\Gamma\left(\frac {8}{14}\right)\frac 2{14}\Gamma\left(\frac {2}{14}\right)\Gamma\left(\frac {18}{14}\right)}\\ =&\pi^{\frac 32}\frac{\frac 8 {14}}{\frac 2 {14}}=4\pi^{\frac 32} \end{align*} and for the denominator, \begin{align*} \Gamma\left(\frac {3}{14}\right)\Gamma\left(\frac {5}{14}\right)\Gamma\left(\frac {13}{14}\right)=& \pi^{\frac 32}\frac{\Gamma\left(\frac {6}{14}\right)\Gamma\left(\frac {10}{14}\right)\Gamma\left(\frac {26}{14}\right)}{\Gamma\left(\frac {10}{14}\right)\Gamma\left(\frac {12}{14}\right)\Gamma\left(\frac {20}{14}\right)}\\ =&\pi^{\frac 32}\frac{\Gamma\left(\frac {6}{14}\right)\Gamma\left(\frac {10}{14}\right)\frac {12}{14}\Gamma\left(\frac {12}{14}\right)}{\Gamma\left(\frac {10}{14}\right)\Gamma\left(\frac {12}{14}\right)\frac{6}{14}\Gamma\left(\frac {6}{14}\right)}\\ =&\pi^{\frac 32}\frac{\frac{12}{14}}{\frac 6{14}}=2\pi^{\frac 32}. \end{align*} Thus it holds $$ \boxed{I = \ln \left(\frac{\Gamma\left(\frac {1}{14}\right)\Gamma\left(\frac {9}{14}\right)\Gamma\left(\frac {11}{14}\right)}{\Gamma\left(\frac {3}{14}\right)\Gamma\left(\frac {5}{14}\right)\Gamma\left(\frac {13}{14}\right)}\right)= \ln 2.} $$

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*}