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 $$\sum_{n=1}^{\infty} \frac{1}{n(2 n+1)}=\sum_{n=1}^{\infty} \frac{1}{n}\int_{0}^{1}x^{2n}dx=\int_{0}^{1}\sum_{n=1}^{\infty} \frac{x^{2n}}{n}=-\int_{0}^{1}\ln(1-x^2)dx $$
Best Answer
One can observe that $$-\log(1-x^k)=\sum_{n\geq 1}\frac{x^{kn}}{n}$$ and integrating this with respect to $x$ on $[0,1]$ we get $$I_{k} =-\int_0^1\log(1-x^k)\,dx=\sum_{n\geq 1}\frac {1}{n(kn+1)}$$ Next we have $$I_k=-\int_{0}^{1}\log(1-x)\,dx-\int_0^1\log\frac{1-x^k}{1-x}\,dx=1-\int_0^1\log(1+x+\dots+x^{k-1})\,dx$$ and we can evaluate the integral for small values of $k$.
For general $k$ we need cyclotomic polynomial and their roots. Let $z_m=\exp(2m i\pi/k) $ for $m=1,2,\dots,k-1$ and then we have $$J_k=\int_0^1\log(1+x+\dots +x^{k-1})\,dx=\sum_{m=1}^{k-1}\int_0^1\log(x-z_m)\,dx$$ If $k$ is even then we have $z_{k/2}=-1$ and rest of the values of $z_m$ can be put into pairs of conjugates and we get $$J_k=\int_0^1\log(1+x)\,dx+\sum_{m<k/2}\int_0^1\log\left(x^2-2x\cos\frac{2m\pi}{k}+1\right)\,dx$$ which equals $$J_k=2\log 2 +1 - k+\sum_{m<k/2}\left\{4\sin^2\left(\frac{m\pi}{k}\right)\log\left(2\sin\left(\frac{m\pi}{k}\right)\right) +\frac{(k-2m)\pi}{k}\cdot\sin\left(\frac{2m\pi}{k}\right)\right\} $$
For odd $k$ all the roots $z_m$ get paired up with their conjugates and the expression for $J_k$ remains same as in case of even $k$ except for the term $2\log 2 $.
We thus have $$I_k=k-2\log 2-\sum_{m<k/2}\left\{4\sin^2\left(\frac{m\pi}{k}\right)\log\left(2\sin\left(\frac{m\pi}{k}\right)\right) +\frac{(k-2m)\pi}{k}\cdot\sin\left(\frac{2m\pi}{k}\right)\right\} $$ if $k$ is even and $$I_k=k- \sum_{m<k/2}\left\{4\sin^2\left(\frac{m\pi}{k}\right)\log\left(2\sin\left(\frac{m\pi}{k}\right)\right) +\frac{(k-2m)\pi}{k}\cdot\sin\left(\frac{2m\pi}{k}\right)\right\} $$ if $k$ is odd.
Using the above formula one can verify the given values in question. As others have noted you have a typo with the sum $I_2$ whose correct value is $2-2\log 2$.
Here is an interesting fact which I noted. Since $I_k>0$ it follows that $$\int_0^1\log(1+x+\dots+x^{k-1})\,dx<1$$ The integrand is positive on $(0,1]$ and if $k$ is large it can take very large value $\log k$ and yet the integral is bounded for all $k$.