Evaluation of $\sum_{n=1}^{\infty}\frac{1}{n(2n+1)}=2-2\ln(2)$ I came across the following statements
$$\sum_{n=1}^{\infty} \frac{1}{n(2 n+1)}=2-2\ln 2 \qquad \tag{1}$$
$$\sum_{n=1}^{\infty} \frac{1}{n(3 n+1)}=3-\frac{3 \ln 3}{2}-\frac{\pi}{2 \sqrt{3}} \qquad \tag{2}$$
$$\sum_{n=1}^{\infty} \frac{1}{n(4 n+1)}=4-\frac{\pi}{2}-3 \ln 2 \qquad \tag{3}$$
$$\sum_{n=1}^{\infty} \frac{1}{n(6 n+1)}=6-\frac{\sqrt{3} \pi}{2}-\frac{3 \ln 3}{2}-2 \ln 2 \qquad \tag{4}$$
The (1) by partial fractions
$$\sum_{n=1}^{\infty}\frac{1}{n(2n+1)}=\sum_{n=1}^{\infty}\frac{1}{n}-\frac{2}{2n+1}$$
$$=\sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+\frac{1}{2}}$$
Recall the Digamma function
$$\psi(x+1)=\gamma+\sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+x}$$
Therefore
$$\sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+\frac{1}{2}}=\psi(1+\frac{1}{2})-\gamma$$
$$\sum_{n=1}^{\infty}\frac{1}{n(2n+1)}=\psi\left(\frac{3}{2}\right)-\gamma$$
In the same token we can derive the relation for the other three ralations. My Question is: can we calculate the values of the digamma function for those values without resorting in the Gauss´s Digamma formula?
$$\psi\left(\frac{r}{m}\right)=-\gamma-\ln (2 m)-\frac{\pi}{2} \cot \left(\frac{r \pi}{m}\right)+2 \sum_{n=1}^{\left\lfloor\frac{m-1}{2}\right\rfloor} \cos \left(\frac{2 \pi n r}{m}\right) \ln \sin \left(\frac{\pi n}{m}\right)$$

I tried this approach also, but I think the resulting integral is divergent $$\sum_{n=1}^{\infty} \frac{1}{n(2 n+1)}=\sum_{n=1}^{\infty} \frac{1}{n}\int_{0}^{1}x^{2n}dx=\int_{0}^{1}\sum_{n=1}^{\infty} \frac{x^{2n}}{n}=-\int_{0}^{1}\ln(1-x^2)dx $$

$\def\o{\omega_k}$
$$
\sum_{n = 1}^\infty  {\frac{1}{{n(2n + 1)}}}  = 2\sum_{n = 1}^\infty\left(\frac1{2n}-\frac1{2n+1}\right)=2\left(1-\sum_{n = 1}^\infty\frac{(-1)^n}n\right)=2-2\log2.
$$
Observe the additional factor $2$ comparing with a typo in your question.

Inspired by this result we can elaborate the general solution:
We note that:
$$S_k=\sum_{n=1}^\infty\frac1{kn(kn+1)}=\sum_{n = 1}^\infty\left(\frac1{kn}-\frac1{kn+1}\right)=1-\frac1k
\sum_{j=1}^k(1-\o^{-j})\log\left(1-\o^j\right),\tag1$$
where
$\o=e^{\frac{2\pi i}k}$
is the $k$-th primitive root of unity.
To verify $(1)$ observe:
$$
\frac1k\sum_{j=1}^k(\o^j)^p=\mathbb{1}_{p\equiv0\pmod k}.
$$
Of course it is of interest to represent the solution (1) in terms of real parameters used in the question. This can be done as follows. Observe that we can omit in the sum the term $j=k$ and combine the complex conjugated terms $j$ and $k-j$ to obtain:
$$
S_k=1-\frac1k\left\{\underbrace{\left[1+(-1)^k\right]}_{s_k}\log2+
\sum_{j=1}^{\left\lfloor\frac {k-1}2\right\rfloor}
\left[(1-\o^{-j})\log\left(1-\o^j\right)+(1-\o^{j})\log\left(1-\o^{-j}\right)\right]\right\},\\
$$
where we singled out the unpaired term $2\log2$ for $j=\frac k2$ in the case of even $k$.
The sum can be rewritten as:
$$\begin{align}
&\sum_{j=1}^{\left\lfloor\frac {k-1}2\right\rfloor}\left\{\left(1-\cos\frac{2\pi j}k\right)\log\left(\left[1-e^{\frac{2\pi j}ki}\right]\left[1-e^{-\frac{2\pi j}ki}\right]\right)+i\sin\frac{2\pi j}k\log\left(
\frac{1-e^{\frac{2\pi j}ki}}{1-e^{-\frac{2\pi j}ki}}\right) \right\}\\
&=\sum_{j=1}^{\left\lfloor\frac {k-1}2\right\rfloor}\left\{\left(1-\cos\frac{2\pi j}k\right)\log\left(2-2\cos\frac{2\pi j}k\right)+\pi\left(1-\frac{2j}k\right)\sin\frac{2\pi j}k \right\},\\
\end{align}$$
so that finally:
$$\begin{align}
S_k&=1-\frac1k\left\{s_k\log2+\sum_{j=1}^{\left\lfloor\frac {k-1}2\right\rfloor}
\left[4\sin^2\frac{\pi j}k \log\left(2\sin\frac{\pi j}k\right)+
\pi\left(1-\frac{2j}k\right)\sin\frac{2\pi j}k\right]\right\}\\
&=1-\log2-\frac1k\left\{4\sum_{j=1}^{\left\lfloor\frac {k-1}2\right\rfloor}\sin^2\frac{\pi j}k \log\left(\sin\frac{\pi j}k\right)+
\frac\pi2\cot\frac\pi k\right\}.\tag2
\end{align}$$
The derivation of the last line is given below.

Appendix
We are going to prove:
$$Z_k=\sum_{0<j<\frac k2}
\left(1-\frac{2j}k\right)\sin\frac{2\pi j}k=\frac12\cot\frac\pi k.$$
First we observe that due to symmetry:
$$
Z_k=\frac12\sum_{0\le j<k}\left(1-\frac{2j}k\right)\sin\frac{2\pi j}k.
$$
Next:
$$\begin{align}
4Z_k\sin\frac\pi k&=\sum_{0\le j<k}\left(1-\frac{2j}k\right)
\left(\cos\frac{\pi(2j-1)}k-\cos\frac{\pi(2j+1)}k\right)\\
&=\sum_{0\le j<k}\left(1-\frac{2j}k\right)\cos\frac{\pi(2j-1)}k
-\sum_{1\le j\le k}\left(1+\frac2k-\frac{2j}k\right)\cos\frac{\pi(2j-1)}k\\
&=\underbrace{\cos\frac{\pi}k}_{j=0}
+\underbrace{\cos\frac{\pi}k}_{j=k}
-\frac2k\underbrace{\sum_{1\le j\le k}\cos\frac{\pi(2j-1)}k}_{=0}\\
&=2\cos\frac{\pi}k.
\end{align}$$
The proof of the another identity used in $(2)$:
$$\sum_{0<j<k}\sin^2\frac{\pi j}k=\frac k2$$
valid for $k\ge2$ is trivial.
