<s>[INST] How to prove that $\int_{0}^{1}{x^2-1\over x^2+1}{x^{2n}\over \ln(x)}{\mathrm dx\over x}=F(n)$ 
$$\int_{0}^{1}{x^2-1\over x^2+1}{x^{2n}\over \ln(x)}{\mathrm dx\over x}=F(n)\tag1$$

$n\ge 1.$

How do we show that $$F(n)=(-1)^{n-1}\ln\left({\pi\over 2}\prod_{k=1}^{n-1}\left({k+1\over k}\right)^{(-1)^k}\right)?$$

$x=\tan(u)$ then $\mathrm dx=\sec^2(u)\mathrm du$
$\cos(2x)={1-\tan^2(x)\over 1+\tan^2(x)}$
$$-\int_{0}^{\pi\over 4}\cos(2u){\tan^{2n}(u)\over \ln(\tan(u))}{\mathrm du\over \cos^2(u)\tan(u)}\tag2$$
$\tan(u)\cos^2(u)={2\over \sin(2u)}$
$$-{1\over 2}\int_{0}^{\pi\over 4}\cot(2u){\tan^{2n}(u)}{\mathrm du\over \ln\tan(u)}\tag3$$
$${\cos(2x)\over \sin(2x)}{\sin^{2n}(x)\over cos^{2n}(x)}=\cos(2x){\sin^{2n-1}(x)\over \cos^{2n+1}(x)}=\tan^{2n-1}(x)-\tan^{2n+1}(x)$$
$$-{1\over 4}\int_{0}^{\pi\over 4}{\tan^{2n-1}(x)\over \ln\tan(x)}\mathrm du+{1\over 4}\int_{0}^{\pi\over 4}{\tan^{2n+1}(x)\over \ln\tan(x)}\mathrm du\tag4$$
 [/INST] For any $s > 0$, we have
\begin{align*}
F(s)
&\stackrel{u=x^2}{=} \int_{0}^{1} \frac{u-1}{u+1} \frac{u^{s-1}}{\log u} \, du
= \int_{0}^{1} \frac{u^{s-1}}{u+1} \left( \int_{0}^{1} u^x \, dx \right) du \\
&= \int_{0}^{1} \int_{0}^{1} \frac{u^{x+s-1}}{u+1} \, dudx
 = \int_{0}^{1} \left( \sum_{k=0}^{\infty} (-1)^k \int_{0}^{1} u^{k+x+s-1} \, du \right) \, dx \\
&= \int_{0}^{1} \left( \sum_{k=0}^{\infty} \frac{(-1)^k}{k+x+s} \right) \, dx
 = \sum_{k=0}^{\infty} (-1)^k \log\left( \frac{k+s+1}{k+s} \right).
\end{align*}
Now for positive integers $n$, we get
\begin{align*}
F(n)
&= (-1)^n \sum_{k=n}^{\infty} (-1)^k \log\left( \frac{k+1}{k} \right) \\
&= (-1)^{n-1} \left( \sum_{k=1}^{n-1} (-1)^k \log\left( \frac{k+1}{k} \right) -
 \sum_{k=1}^{\infty} (-1)^k \log\left( \frac{k+1}{k} \right) \right).
\end{align*}
Finally the last infinite sum can be computed by Wallis product, hence the identity follows.
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