$\int_{0}^{1}{2n-x-x^3-x^5-\cdots-x^{4n-1}\over 1+x^2}\cdot{\mathrm dx\over \ln{x}}$ Consider

$$\int_{0}^{1}{2n-x-x^3-x^5-\cdots-x^{4n-1}\over 1+x^2}\cdot{\mathrm dx\over \ln{x}}=I\tag1$$
  $n\ge1$
How does one show that $I=2n\ln{\Gamma(3/4)\over \Gamma(5/4)}-\ln{[8^n(2n-1)!!]}?$

An attempt:
$J=x+x^3+x^5+x^7+\cdots+$
$J=x(1+x^2+x^4+x^6+\cdots+)$
Geometric series
$1+x+x^2+x^3+\cdots x^{n-1}={x(1-x^n)\over 1-x}$
$1+x^2+x^4+x^6+\cdots+x^{2n-2}={x^2(1-x^{2n})\over 1-x^2}$
$I$ becomes
$$2n\int_{0}^{1}{1\over 1+x^2}\cdot{\mathrm dx\over \ln{x}}-\int_{0}^{1}{1-x^{2n}\over 1+x^2}\cdot{x^3\over \ln{x}}\mathrm dx=I\tag2$$
$x=\tan{y}$ then $dx=\sec^2{y}dy$
$$\int_{0}^{\pi/4}{1\over \ln{\tan{y}}}\mathrm dy-\int_{0}^{\pi/4}{1-\tan^{2n}{y}\over \ln{\tan{y}}}\cdot\tan^3{y}\mathrm dy\tag3$$
$$\int_{0}^{\pi/4}{1-\tan^3{y}\over \ln{\tan{y}}}\mathrm dy-\int_{0}^{\pi/4}{\tan^{2n}{y}\over \ln{\tan{y}}}\cdot\tan^3{y}\mathrm dy\tag4$$
$1-x^3=(1-x)(1+x+x^2)$
I am not sure what to do next
 A: The first term in the LHS of your $(2)$ is a divergent integral: better to avoid them.
Instead, we may consider that:
$$ I(k)=\int_{0}^{1}\frac{1-x^{2k-1}}{1+x^2}\cdot\frac{dx}{\log(x)}=-\int_{0}^{+\infty}\frac{e^{-x}-e^{-2kx}}{1+e^{-2x}}\cdot\frac{dx}{x} $$
can be easily dealt with through a geometric series expansion and Frullani's theorem, leading to:
$$ I(k) = \log\frac{2k}{1}-\log\frac{2k+2}{3}+\log\frac{2k+4}{5}-\log\frac{2k+6}{7}+\ldots $$
then to:
$$ I(k) = -\log\left(\frac{2k}{1}\cdot\frac{3}{2k+2}\cdot\frac{2k+4}{5}\cdot\frac{7}{2k+6}\cdots\right) $$
and finally to:
$$ I(k) = (-1)^{k+1}\log\sqrt{\pi} +\log\left(\frac{(k-2)!!\,\Gamma\left(\frac{3}{4}\right)}{2(k-1)!!\,\Gamma\left(\frac{5}{4}\right)}\right)$$
that is simple to sum over $k$.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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Basically, the whole problem is reduced to evaluate the following integral:

\begin{align}
&\int_{0}^{1}{1 - x^{\mu} \over 1 + x^{2}}\,{\dd x \over \ln\pars{x}}  =
\int_{0}^{1}{1 - x^{2} - x^{\mu} + x^{\mu + 2}\over 1 - x^{4}}
\,{\dd x \over \ln\pars{x}}
\\[5mm] \stackrel{x^{4}\ \mapsto\ x}{=} &
\int_{0}^{1}{x^{-3/4} - x^{-1/4} - x^{\mu/4 - 3/4} + x^{\mu/4 - 1/4} \over
1 - x}\,{\dd x \over \ln\pars{x}}
\\[5mm] = &\
\int_{0}^{1}{x^{-3/4} - x^{-1/4} - x^{\mu/4 - 3/4} + x^{\mu/4 - 1/4} \over
1 - x}\pars{-\int_{0}^{\infty}x^{t}\,\dd t}\,\dd x
\\[5mm] = &\
\int_{0}^{\infty}\int_{0}^{1}
{-x^{t - 3/4} + x^{t - 1/4} + x^{t + \mu/4 - 3/4} - x^{t + \mu/4 - 1/4} \over
1 - x}\,\dd x\,\dd t
\\[5mm] = &\
\int_{0}^{\infty}\bracks{\Psi\pars{t + {1 \over 4}} -
\Psi\pars{t + {3 \over 4}} - \Psi\pars{t + {\mu + 1\over 4}} +
\Psi\pars{t + {\mu + 3 \over 4}}}\dd t
\\[5mm] = &\
\left.
\ln\pars{\Gamma\pars{t + 1/4}\Gamma\pars{t + \bracks{\mu + 3}/4} \over
\Gamma\pars{t + 3/4}\Gamma\pars{t + \bracks{\mu + 1}/4}}
\right\vert_{\ t\ =\ 0}^{\ t\ \to\ \infty} =
\bbx{\ds{-\ln\pars{{\root{2} \over 2\pi}\,\Gamma^{2}\pars{1 \over 4}\,
{\Gamma\pars{\bracks{\mu + 3}/4} \over \Gamma\pars{\bracks{\mu + 1}/4}}}}}
\end{align}
