<s>[INST] $\int_{0}^{1}{2n-x-x^3-x^5-\cdots-x^{4n-1}\over 1+x^2}\cdot{\mathrm dx\over \ln{x}}$ 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
 [/INST] 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 $k$.
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