Need help with $\int_0^1\frac{\log(1+x)-\log(1-x)}{\left(1+\log^2x\right)x}\,dx$

Recall the Frullani Integral: $$ \int_0^\infty\frac{e^{-ax}-e^{-bx}}{x}\,\mathrm{d}x =\log(b/a)\tag{1} $$ and scale equation $(4)$ from this answer to get $$ \sum_{k=0}^\infty\frac1{(2k+1)^2+u^2} =\frac\pi{4u}\tanh\left(\frac{\pi u}2\right)\tag{2} $$ Then $$ \begin{align} &\int_0^1\frac{\log(1+x)-\log(1-x)}{\left(1+\log^2(x)\right)x}\,\mathrm{d}x\\ &=\int_0^\infty\frac{\log\left(1+e^{-u}\right)-\log\left(1-e^{-u}\right)}{1+u^2}\,\mathrm{d}u\tag{3a}\\ &=2\sum_{k=0}^\infty\int_0^\infty\frac{e^{-(2k+1)u}}{2k+1}\frac{\mathrm{d}u}{1+u^2}\tag{3b}\\ &=2\sum_{k=0}^\infty\int_0^\infty\frac{e^{-u}\,\mathrm{d}u}{(2k+1)^2+u^2}\tag{3c}\\ &=2\int_0^\infty\frac\pi{4u}\tanh\left(\frac{\pi u}2\right)e^{-u}\,\mathrm{d}u\tag{3d}\\ &=\frac\pi2\int_0^\infty\frac{1-e^{-\pi u}}{u}\frac{e^{-u}}{1+e^{-\pi u}}\,\mathrm{d}u\tag{3e}\\ &=\frac\pi2\int_0^\infty\sum_{k=0}^\infty(-1)^k\frac{e^{-(k\pi+1)u}-e^{-((k+1)\pi+1)u}}u\,\mathrm{d}u\tag{3f}\\ &=\frac\pi2\sum_{k=0}^\infty(-1)^k\log\left(\frac{(k+1)\pi+1}{k\pi+1}\right)\tag{3g}\\ &=\frac\pi2\sum_{k=0}^\infty\log\left(\frac{(2k+1)\pi+1}{2k\pi+1}\frac{(2k+1)\pi+1}{(2k+2)\pi+1}\right)\tag{3h}\\ &=\lim_{n\to\infty}\frac\pi2\log\left[\prod_{k=0}^n\frac{k+\frac12+\frac1{2\pi}}{k+\frac1{2\pi}}\frac{k+\frac12+\frac1{2\pi}}{k+1+\frac1{2\pi}}\right]\tag{3i}\\ &=\lim_{n\to\infty}\frac\pi2\log\,\left[\frac{\Gamma\left(n+\frac32+\frac1{2\pi}\right)^2}{\Gamma\left(\frac12+\frac1{2\pi}\right)^2}\frac{\Gamma\left(\frac1{2\pi}\right)\Gamma\left(1+\frac1{2\pi}\right)}{\Gamma\left(n+1+\frac1{2\pi}\right)\Gamma\left(n+2+\frac1{2\pi}\right)}\right]\tag{3j}\\ &=\frac\pi2\log\,\left[\frac{\Gamma\left(\frac1{2\pi}\right)\Gamma\left(1+\frac1{2\pi}\right)}{\Gamma\left(\frac12+\frac1{2\pi}\right)^2}\right]\tag{3k}\\ &=\bbox[5px,border:2px solid #C0A000]{\pi\log\,\left[\frac{\frac1{\sqrt{2\pi}}\Gamma\left(\frac1{2\pi}\right)}{\Gamma\left(\frac12+\frac1{2\pi}\right)}\right]}\tag{3m} \end{align} $$ Explanation:
$\text{(3a)}$: Substitute $x=e^{-u}$
$\text{(3b)}$: $\log\left(\frac{1+x}{1-x}\right)=2\sum\limits_{k=0}^\infty\frac{x^{2k+1}}{2k+1}$
$\text{(3c)}$: Substitute $u\mapsto\frac u{2k+1}$
$\text{(3d)}$: apply $(2)$
$\text{(3e)}$: $\tanh\left(\frac{\pi u}2\right)=\frac{1-e^{-\pi u}}{1+e^{-\pi u}}$
$\text{(3f)}$: $\frac1{1+x}=\sum\limits_{k=0}^\infty(-1)^kx^k$
$\text{(3g)}$: apply $(1)$
$\text{(3h)}$: combine $2k$ and $2k+1$ terms
$\text{(3i)}$: change a sum of logs to a log of a product
$\text{(3j)}$: write products as ratios of Gamma functions
$\text{(3k)}$: apply Gautschi's Inequality
$\text{(3m)}$: $\Gamma(1+x)=x\Gamma(x)$

An alternative way to evaluate $$\frac{\pi}{2} \int_{0}^{\infty} \tanh \left(\frac{\pi u}{2} \right) \frac{e^{-u}}{u} \, du ,$$ which is line $3d$ in robjohn's answer, is to add a parameter and then differentiate under the integral sign.

Specifically, let $$I(a) = \frac{\pi}{2}\int_{0}^{\infty} \tanh \left(\frac{\pi u}{2} \right) \frac{e^{-au}}{u} \, du.$$

Then $$ \begin{align} I'(a) &= - \frac{\pi}{2} \int_{0}^{\infty} \tanh \left(\frac{\pi u}{2} \right) e^{-au} \, du \\ &= -\frac{\pi}{2} \int_{0}^{\infty} \left(\frac{1}{1+e^{- \pi u}}- \frac{e^{- \pi u}}{1+e^{-\pi u}} \right)e^{-au} \, du \\ &= -\frac{\pi}{2} \int_{0}^{\infty} \left(\sum_{n=0}^{\infty} (-1)^{n} e^{-n \pi u} + \sum_{n=1}^{\infty} (-1)^{n} e^{-n \pi u} \right)e^{-au} \, du \\ &= \frac{\pi}{2} \int_{0}^{\infty} \left(1- 2 \sum_{n=0}^{\infty} (-1)^{n}e^{-n \pi u} \right) e^{-au} \, du \\ &= \frac{\pi }{2a} -\pi \sum_{n=0}^{\infty} \frac{(-1)^{n}}{a+n \pi} \\ &= \frac{\pi }{2a} -\frac{1}{2} \psi \left(\frac{a+\pi}{2 \pi} \right) + \frac{1}{2} \psi \left(\frac{a}{2 \pi} \right) \tag{1}. \end{align}$$

Integrating back, we get $$ \begin{align} I(a) &= \frac{\pi}{2} \log(a) - \pi \log \Gamma \left(\frac{a+\pi}{2 \pi} \right) + \pi \log \Gamma\left(\frac{a}{2 \pi} \right) +C \\ &= \pi \log \left(\frac{\sqrt{a} \, \Gamma \left(\frac{a}{2 \pi } \right)}{\Gamma \left(\frac{a}{2 \pi} + \frac{1}{2} \right)} \right) + C,\end{align} $$

where $$\lim_{a \to \infty} I(a) =0 = \lim_{a \to \infty} \pi \log \left(\frac{\sqrt{a} \, \Gamma \left(\frac{a}{2 \pi } \right)}{\Gamma \left(\frac{a}{2 \pi} + \frac{1}{2} \right)} \right) +C $$

$$= \pi \log (\sqrt{2 \pi}) + C. \tag{2} $$

Therefore,

$$\frac{\pi}{2} \int_{0}^{\infty} \tanh \left(\frac{\pi u}{2} \right) \frac{e^{-u}}{u} \, du = I(1) =\pi \log \left(\frac{ \Gamma \left(\frac{1}{2 \pi } \right)}{\sqrt{2 \pi} \, \Gamma \left(\frac{1}{2 \pi} + \frac{1}{2} \right)} \right).$$

$$ $$

$(1)$ http://mathworld.wolfram.com/DigammaFunction.html (6)

$(2)$ In general, for $x,y>0$, $\lim_{a \to \infty} \frac{a^{x} \Gamma(ya)}{\Gamma(ya+x)} = y^{-x}$. This can be proven using Stirling's approximation formula for the gamma function.