The following article from Wikipedia show us these formulas for harmonic numbers $H_x$: the Taylor series for $|x|<1$, and formulas for $\int_0^m H_x dx$ for integers $m\geq 1$.
Now one defines for real numbers $0<x<1$ $$M_x:=\sum_{n=2}^\infty (-1)^n \frac{x^{n-1}}{\zeta(n)}.\tag{1}$$ (I don't know if previous real numbers $M_x$ are well-known.)
Question. A) Can you calculate or provide me hints to calculate a good approximation of $$\int_0^1 M_x dx?$$
B) Do you know if was in the literature a closed-form, for each fixed integer $m>1$, of $$\int_0^m M_x dx$$ in terms of special functions, or well asymptotics of such definite integral as $m\to\infty$? If yes, then aren't required more details, please refer the literature. Thanks in advance.
$$M_x = \sum_{n\geq 2}(-1)^n x^{n-1}\sum_{m\geq 1}\frac{\mu(m)}{m^n} = x\sum_{m\geq 1}\frac{\mu(m)}{m(m+x)}=-\sum_{m\geq 1}\frac{\mu(m)}{m+x}\tag{1}$$ hence $$\int_{0}^{1}M_x\,dx = -\sum_{m\geq 1}\mu(m)\log\left(1+\frac{1}{m}\right)=\color{red}{\sum_{n\geq 2}\frac{(-1)^n}{n\,\zeta(n)}}\approx 0.15294. \tag{2}$$ The integral over a different interval can be managed in a similar way, leading to a slow-convergent series that can be easily accelerated, for instance by approximating $\frac{1}{\zeta(n)}$ with $1-\frac{1}{2^n}$ for large values of $n$.