This question has an interesting aspect that merits being commented, which is that we can compute a closed form for the sum using Mellin transforms and harmonic sums.
Introduce the sum $S(x)$ given by
$$S(x) = \sum_{n\ge 1} \left(1- xn \log\frac{2xn+1}{2xn-1}\right)$$
so that we are interested in $S(1).$
As mentioned before the sum term is harmonic and may be evaluated by inverting its Mellin transform.
Recall the harmonic sum identity
$$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) =
\left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$
where $g^*(s)$ is the Mellin transform of $g(x).$
In the present case we have
$$\lambda_k = 1, \quad \mu_k = k \quad \text{and} \quad
g(x) = 1-x\log\frac{2x+1}{2x-1} =
1-x\log\left(1+\frac{2}{2x-1}\right).$$
The abscissa of convergence of $$\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} = \zeta(s)$$
is $\Re(s)>1.$
Now to find the Mellin transform $g^*(s)$ of $g(x)$ the fundamental strip being $\langle 0,2\rangle$ (which contains the abscissa of convergence of the Dirichlet series sum term) we first use integration by parts to get
$$\int_0^\infty \left(-x\log\left(1+\frac{2}{2x-1}\right)\right) x^{s-1} dx
\\= \left[\left(-x\log\left(1+\frac{2}{2x-1}\right)\right)
\frac{x^{s+1}}{s+1}\right]_0^\infty
- \int_0^\infty \frac{4}{4x^2-1} \frac{x^{s+1}}{s+1} dx
= - \int_0^\infty \frac{4}{4x^2-1} \frac{x^{s+1}}{s+1} dx.$$
with the fundamental strip being $\langle -1, 0\rangle$.
It becomes evident that we require the following Mellin transform:
$$h^*(s) = \int_0^\infty h(x) x^{s-1} dx$$
where $$h(x) = \frac{4}{4x^2-1}.$$
This transform integral is not strictly speaking convergent but we can compute its principal value by using a semicircular contour in the upper half plane that is traversed clockwise and picks up half the residues at the two poles at $\pm 1/2.$ This gives
$$h^*(s) (1-e^{i\pi s})
= \frac{1}{2} \times 2 \pi i
\left(\mathrm{Res}(h(x) x^{s-1}; x=1/2) + \mathrm{Res}(h(x) x^{s-1}; x=-1/2)\right)$$
which gives
$$h^*(s) (1-e^{i\pi s}) = \frac{1}{2} \times 2 \pi i ((1/2)^{s-1}-(-1/2)^{s-1})
= 2^{-s} \times 2 \pi i (1+(-1)^s).$$
This yields
$$h^*(s) = 2^{-s} \times 2 \pi i \frac{1+e^{i\pi s}}{1-e^{i\pi s}}
= 2^{-s} \times 2 \pi i \frac{e^{-i\pi s/2}+e^{i\pi s/2}}{e^{-i\pi s/2}-e^{i\pi s/2}}
= -\frac{2}{2^s} \pi \cot(\pi s/2).$$
Returning to $g^*(s)$ we have
$$g^*(s) = \frac{2}{2^{s+2}} \frac{1}{s+1} \pi \cot(\pi (s+2)/2)
= \frac{1}{2^{s+1}} \frac{1}{s+1} \pi \cot(\pi s / 2).$$
Therefore the Mellin transform $Q(s)$ of $S(x)$ is given by
$$\frac{1}{2^{s+1}} \frac{1}{s+1} \pi \cot(\pi s / 2)\zeta(s).$$
The Mellin inversion integral here is
$$\frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} Q(s)/x^s ds$$
which we evaluate by shifting it to the right for an expansion about infinity.
We must sum the residues at the poles at the positive even integers of $Q(s)/x^s$ which are
$$\mathrm{Res}(Q(s)/x^s; s=2q) =
\frac{1}{2^{2q+1}} \frac{1}{2q+1} \times 2 \times \zeta(2q) \frac{1}{x^{2q}}
\\ =\frac{1}{2^{2q+1}} \frac{1}{2q+1} \times 2 \times
\frac{(-1)^{q+1} B_{2q} (2\pi)^{2q}}{2\times (2q)!} \frac{1}{x^{2q}}
= \frac{1}{2} \times \frac{(-1)^{q+1} B_{2q} \pi^{2q}}{(2q+1)!} \frac{1}{x^{2q}}.$$
We thus require the sum
$$ - \frac{1}{2} \sum_{q\ge 1}
\frac{i^{2q} B_{2q} \pi^{2q}}{(2q+1)!} \frac{1}{x^{2q}}.$$
The exponential generating function of the even Bernoulli numbers is
$$-1 + \frac{1}{2} t + \frac{t}{e^t-1} = \sum_{q\ge 1} B_{2q} \frac{t^{2q}}{(2q)!}.$$
Integrate this to obtain
$$-t - \frac{t^2}{4} + t\log(1-e^t) + \mathrm{Li}_2(e^t).$$
The constant that appeared during the integration was the well-known zeta function value $\mathrm{Li}_2(1) = \zeta(2) = \pi^2/6$ so that we finally have
$$ \sum_{q\ge 1} B_{2q} \frac{t^{2q}}{(2q+1)!}
= \frac{1}{t}
\left(-\frac{\pi^2}{6}-t - \frac{t^2}{4} + t\log(1-e^t) + \mathrm{Li}_2(e^t)\right).$$
To conclude use another well known zeta function value which is $\mathrm{Li}_2(-1) = -\pi^2/12$ (derived from $\mathrm{Li}_2(1)$) and put $t=i\pi /x = i\pi$ to obtain that (we lose the minus sign because we are shifting to the right)
$$S(1) = \frac{1}{2} \frac{1}{i\pi}
\left(-\frac{\pi^2}{6}-i\pi + \frac{\pi^2}{4} + i\pi\log 2 + \mathrm{Li}_2(-1)\right)
\\ = \frac{1}{2} \frac{1}{i\pi} \left(-i\pi + i\pi \log 2 \right)
= \frac{1}{2} (\log 2 - 1).$$
Remark. The integral of the generating function of the Bernoulli numbers is easily verified by differentiation:
$$\left(-\frac{t^2}{2} + t\log(1-e^t) + \mathrm{Li}_2(e^t)\right)'
= -t + \frac{t}{1-e^t} (-e^t) + \log(1-e^t) +
\left(\sum_{n\ge 1} \frac{e^{nt}}{n^2}\right)'
\\ = \frac{-t+te^t}{1-e^t} - \frac{te^t}{1-e^t}
+ \log(1-e^t) + \sum_{n\ge 1} \frac{e^{tn}}{n}
\\ = \frac{-t}{1-e^t} + \log(1-e^t) - \log(1-e^t)
= \frac{t}{e^t-1}.$$