Prove that $\int_{0}^{1}{(1-x)(x-3)\over 1+x^2}\cdot{dx\over \ln{x}}={\ln{8\over \Gamma^4(3/4)}}$

Hint. One may set $$ f(s):=\int_0^1 \frac{x^{4s}-1}{(1+x^2)\ln x}dx, \quad s>0. \tag1 $$ In order to get rid of the factor $\ln x$ in the denominator, we may differentiate under the integral sign getting $$ f'(s)=4\int_0^1 \frac{x^{4s}}{1+x^2}dx=4\int_0^1 \frac{x^{4s}(1-x^2)}{1-x^4}dx, \quad s>0, \tag2 $$ giving $$ \begin{align} f'(s)&=\psi\left(s+\frac34\right)-\psi\left(s+\frac14\right),\tag3 \end{align} $$ where we have used the standard integral representation of the digamma function $$ \int_{0}^{1}{1 - t^{s - 1} \over 1 - t}\,dt \, = \psi (s)+ \gamma, \quad s>0, $$ $\gamma$ being the Euler-Mascheroni constant.

Then integrating $(3)$, observing that as $s \to 0^+$, $f(s) \to 0$, one gets

$$ f(s)=\int_0^1 \frac{x^{4s}-1}{(1+x^2)\ln x}dx=\log\left(\frac{\Gamma\left(\frac14\right)\Gamma\left(s+\frac34\right)}{\Gamma\left(\frac34\right)\Gamma\left(s+\frac14\right)}\right), \quad s>0, \tag4 $$

from which you deduce the value of your initial integral by writing $$ \int_{0}^{1}{(1-x)(x-3)\over 1+x^2}\cdot{dx\over \ln{x}}=f(1/2)+4f(1/4). $$