Harmonic series with sign alternates every $n$ terms. Let $A(1)=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots$
Let $A(2)=\frac{1}{1}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}+\dots$
Let $A(3)=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}+\dots$
In general, for $n \in \mathbb{N}$,  let $A(n)$ be the same series as the above with sign alternates every $n$ terms.
I know that $A(1)=\log(2)$, and $A(2)=\frac{1}{4}(\pi+\log(4))$.
How can we evaluate $A(n)$ for $n \in \mathbb{N_{\ge3}}$?

$A_n$ can be expressed as
$$
A_n=\sum_{k=1}^n \sum_{u=0}^\infty \frac{(-1)^u}{nu+k}.
$$
Now, from the Taylor series of $\frac{1}{1+x}$ for $|x|<1$ we have
$$
{
\frac{1}{1+x}=\sum_{u=0}^\infty (-1)^ux^u\implies
\\
\frac{1}{1+x^n}=\sum_{u=0}^\infty (-1)^ux^{nu}\implies
\\
\frac{x^{k-1}}{1+x^n}=\sum_{u=0}^\infty (-1)^ux^{nu+k-1}\implies
\\
\int_0^1\frac{x^{k-1}}{1+x^n}dx=\sum_{u=0}^\infty \frac{(-1)^u}{nu+k}
}
$$
hence
$$
A_n{=\sum_{k=1}^n \int_0^1\frac{x^{k-1}}{1+x^n}dx
\\=
\int_0^1\sum_{k=1}^n\frac{x^{k-1}}{1+x^n}dx
\\=
\int_0^1\frac{1-x^n}{1+x^n}\frac{1}{1-x}dx
\\=
\int_0^1\sum_{k=1}^n \frac{2u_k}{u_k-1}\frac{1}{x-u_k} dx
\\=
\sum_{k=1}^n\frac{2u_k}{u_k-1}\int_0^1 \frac{1}{x-u_k} dx
\\=
\sum_{k=1}^n\frac{2u_k}{u_k-1}[\ln(1-u_k)-\ln(-u_k)]
}
$$
where $u_k=\exp(i\pi\frac{2k+1}{n})$ is the $k$th root of $-1$ and $\ln$ is the main branch of complex logarithm.
Additional Remarks
The reason for interchanging integrals and summations, is the existence and boundedness of the integral values and the boundedness of summation scope.
