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

$$I=\int_{0}^{1}{(1-x)(x-3)\over 1+x^2}\cdot{dx\over \ln{x}}=\color{blue}{\ln{8\over \Gamma^4(3/4)}}\tag1$$

$(1-x)(x-3)=-x^2+4x-3$
$${1\over 1+x^2}=\sum_{n=0}^{\infty}(-1)^nx^{2n}\tag2$$
$$I=-\sum_{n=0}^{\infty}(-1)^n\int_{0}^{\infty}{x^{2n+2}-4x^{2n+1}+3x^{2n}\over \ln{x}}dx\tag3$$
Rewrite (3) to apply Frullani's theorem
$$I=-\sum_{n=0}^{\infty}(-1)^n\int_{0}^{\infty}{x^{2n+2}-x^{2n+1}-3x^{2n+1}+3x^{2n}\over \ln{x}}dx\tag4$$
$$I=\sum_{n=0}^{\infty}(-1)^{n-1}\ln\left({2n+3\over (2n+2)^4}\cdot{(2n+1)^3}\right)\tag5$$
This method is a bit boring method! I converted (1) into series and using Frullani's theorem and again have to solve the series is another step before we can reached our answer.
How can I solve (1) without using series?
 A: 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 
$$
\int_{0}^{1}{(1-x)(x-3)\over 1+x^2}\cdot{dx\over \ln{x}}=f(1/2)+4f(1/4).
$$
