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 $k$.
As I mentioned in my question, when integrating back from $0$ to $z$ the L.H.S. is not our goal integral, nevertheless, we can still evaluate the integral by this method. Here is how to do so:
First, lets integrate back the first integral, namely:
$$
\begin{aligned}
&\int_0^z\,\int_0^\infty\frac{2wx }{(w^2+x^2)(e^{2\pi x}-1)}\,dx\,dw=\int_0^zw\log(w)\,dw-\int_0^zw\psi(w)\,dw-\frac12\int_0^zdw\\
&\int_0^\infty\frac{x \ln(z^2+x^2)}{e^{2\pi x}-1}\,dx-2\int_0^\infty\frac{x \ln(x)}{e^{2\pi x}-1}\,dx=\frac{z^2 \ln z}{2}-\frac{z^2}{4}-z \ln \Gamma(z)+\int_0^z\ln \Gamma(w)dw-\frac{z}{2}\\
&\int_0^\infty\frac{x \ln(z^2+x^2)}{e^{2\pi x}-1}\,dx=2\int_0^\infty\frac{x \ln(x)}{e^{2\pi x}-1}\,dx+\frac{z^2 \ln z}{2}-\frac{z^2}{4}-z \ln \Gamma(z)+\int_0^z\ln \Gamma(w)dw-\frac{z}{2}\\
&\int_0^\infty\frac{x \ln(z^2+x^2)}{e^{2\pi x}-1}\,dx=\zeta^{\prime}(-1)+\frac{z^2 \ln z}{2}-\frac{z^2}{4}-z \ln \Gamma(z)+\int_0^z\ln \Gamma(w)dw-\frac{z}{2}\\
\end{aligned}
$$
Taking the limit $z \to 1$
$$\int_0^\infty\frac{x \ln(1+x^2)}{e^{2\pi x}-1}\,dx=\zeta^{\prime}(-1)-\frac{3}{4}+\frac12 \ln 2 \pi \qquad \blacksquare$$
Integrating back the second integral:
$$\begin{aligned}
&\int_0^z\,\int_0^\infty\frac{2wx }{(w^2+x^2)(e^{\pi x}-1)}\,dx\,dw=\int_0^zw\log\left(\frac{w}{2}\right)\,dw-\int_0^z w\psi\left(\frac{w}{2}\right)\,dw-\int_0^z dw\\
&\int_0^\infty\frac{x \ln(z^2+x^2) }{(e^{\pi x}-1)}\,dx=2\int_0^\infty\frac{x \ln(x) }{(e^{\pi x}-1)}\,dx+\frac{z^2\ln z}{2}-\frac{z^2}{4}-\frac{z^2 \ln 2 }{2}-z-4\int_0^{z/2} w\psi\left(w\right)\,dw\\
&\int_0^\infty\frac{x \ln(z^2+x^2) }{(e^{\pi x}-1)}\,dx=2\int_0^\infty\frac{x \ln(x) }{(e^{\pi x}-1)}\,dx+\frac{z^2\ln z}{2}-\frac{z^2}{4}-\frac{z^2 \ln 2 }{2}-z-4\left(\frac{z}{2} \ln\Gamma\left(\frac{z}{2}\right) -\int_0^{z/2} \ln\Gamma\left(w\right)\,dw \right)\\
\end{aligned}
$$
Letting $z \to 1$
$$
\begin{aligned}
\int_0^\infty\frac{x \ln(1+x^2) }{(e^{\pi x}-1)}\,dx&=2\left(\frac16+\frac{\ln 2}{6}-2 \ln A \right)-\frac{1}{4}-\frac{ \ln 2 }{2}-1-4\left(\frac{1}{2} \ln\Gamma\left(\frac{1}{2}\right) -\int_0^{1/2} \ln\Gamma\left(w\right)\,dw \right)\\
&=\frac13-4 \ln A -\frac{5}{4}-\frac{ \ln 2 }{6}-4\left(\frac{1}{4} \ln \pi -\left( \frac32\ln A+\frac{5}{24}\ln 2+\frac14 \ln \pi\right)\right)\\
&=\frac13-4 \ln A -\frac{5}{4}-\frac{ \ln 2 }{6}+6\ln A+\frac56\ln 2\\
&=2\ln A -\frac{11}{12}+\frac{ 2 }{3}\ln 2\\
&=-2\zeta^{\prime}(-1)+\frac{ 2 }{3}\ln 2-\frac34 \qquad \blacksquare\\
\end{aligned}
$$
We the conclude
$$
\begin{aligned}
&\int_{0}^{\infty} \frac{x \ln \left(1+x^{2}\right)}{e^{2 \pi x}-1} d x=\zeta^{\prime}(-1)+\ln \sqrt{2 \pi}-\frac{3}{4} \\
& \\
&\int_{0}^{\infty} \frac{x \ln \left(1+x^{2}\right)}{e^{\pi x}-1} d x=-2 \zeta^{\prime}(-1)+\frac{2}{3} \ln 2-\frac{3}{4}
\end{aligned}
$$
Therefore our integral equals
$$
\begin{aligned}
\int_0^\infty\frac{x \ln(1+x^2)}{\sinh \pi x}\,dx&=2\left(-2 \zeta^{\prime}(-1)+\frac{2}{3} \ln 2-\frac{3}{4}-\zeta^{\prime}(-1)-\ln \sqrt{2 \pi}+\frac{3}{4}\right)\\
&=2\left(-3 \zeta^{\prime}(-1)+\frac{2}{3}\ln 2-\frac12 \ln 2-\frac12 \ln \pi\right)\\
&=-6 \zeta^{\prime}(-1)+\frac{4}{3}\ln 2- \ln 2- \ln \pi\\
&=-6 \zeta^{\prime}(-1)+\frac{\ln 2}{3}- \ln \pi\\
&=-6 \left(\frac{1}{12}-\ln A \right)+\frac{\ln 2}{3}- \ln \pi\\
&=-\frac{\ln e}{2}+\ln A^6 +\frac{\ln 2}{3}- \ln \pi\\
&=\ln \, \frac{\sqrt[3]{2}\, A^6}{\pi \sqrt{e}} \qquad \blacksquare
\end{aligned}
$$
Appendix
Consider the Integral
$$
\begin{aligned}
I(s)&=\int_0^\infty \frac{x^{s-1}}{e^{cx}-1}\,dx\\
&=\frac{1}{c^s}\int_0^\infty \frac{x^{s-1}}{e^{x}-1}\,dx\\
&=\frac{1}{c^s}\int_0^\infty \frac{e^{-x}x^{s-1}}{1-e^{-x}}\,dx\\
&=\frac{1}{c^s}\sum_{k=1}^\infty \, \int_0^\infty x^{s-1}e^{-kx}\,dx\\
&=\frac{1}{c^s}\sum_{k=1}^\infty \frac{1}{k^s} \, \int_0^\infty x^{s-1}e^{-x}\,dx\\
&=\frac{1}{c^s}\sum_{k=1}^\infty \frac{1}{k^s} \, \int_0^\infty x^{s-1}e^{-x}\,dx\\
&=\frac{\Gamma(s)\zeta(s)}{c^s}
\end{aligned}
$$
Differentiating both sides w.r. to s
$$
\begin{aligned}
\int_0^\infty \frac{x^{s-1} \ln x}{e^{cx}-1}\,dx&=\frac{d}{ds}\left(\frac{\Gamma(s)\zeta(s)}{c^s} \right)\\
&=\frac{c^s\left(\Gamma^{\prime}(s)\zeta(s)+\Gamma(s)\zeta^{\prime}(s) \right)-c^s\ln c \left( \Gamma(s)\zeta(s) \right)}{(c^s)^2}\\
&=\frac{\left(\Gamma(s)\psi(s)\zeta(s)+\Gamma(s)\zeta^{\prime}(s) \right)-\ln c \left( \Gamma(s)\zeta(s) \right)}{c^s}\\
\end{aligned}
$$
Setting $s=2$
$$
\begin{aligned}
\int_0^\infty \frac{x \ln x}{e^{cx}-1}\,dx&=\frac{\left(\Gamma(2)\psi(2)\zeta(2)+\Gamma(2)\zeta^{\prime}(2) \right)-\ln c \left( \Gamma(2)\zeta(2) \right)}{c^2}\\
&=\frac{\left((1-\gamma)\zeta(2)+\zeta(2)\left(\ln 2\pi+\gamma-12 \ln A\right) \right)-\zeta(2)\ln c }{c^2}\\
&=\zeta(2)\left(\frac{1-\ln c+\ln 2\pi-12 \ln A }{c^2}\right)\\
\end{aligned}
$$
Setting $c=\pi$
$$\int_0^\infty \frac{x \ln x}{e^{\pi x}-1}\,dx=\frac16+\frac{\ln 2}{6}-2 \ln A \qquad \blacksquare$$
For the LogGamma integral between $0$ and $1/2$
Recall Kummer´s fourier expansion for LogGamma $0<x<1$
$$\ln\left(\Gamma(x)\right)=\frac{\ln\left(2 \pi\right)}{2}+ \sum_{k=1}^{\infty}\frac{1}{2k}\cos\left(2 \pi k x\right) + \sum_{k=1}^{\infty}\frac{\gamma+\ln\left(2 \pi k\right)}{ \pi k}\sin\left(2 \pi k x \right) \tag{A.1}$$
Integrating we obtain
$$\int_0^x\ln\left(\Gamma(u)\right)du=\frac{x \ln\left(2 \pi\right)}{2}+\frac{1}{4 \pi}\sum_{k=1}^{\infty}\frac{\sin\left(2 \pi k x \right)}{k^2} +\frac{\gamma}{2 \pi^2}\sum_{k=1}^{\infty}\frac{1}{k^2}-\frac{\gamma}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\cos \left(2 \pi k x \right)}{k^2}+\frac{1}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\ln \left( 2 \pi k\right)}{k^2}-\frac{1}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\ln \left( 2 \pi k\right) \cos\left(2 \pi k x \right)}{k^2} \tag{A.2}$$
Letting $x=\frac12$ in $(A.2)$ we obtain
$$\begin{aligned}
\int_0^{1/2}\ln\left(\Gamma(x)\right)dx&=\frac{ \ln\left(2 \pi\right)}{4} +\frac{\gamma}{2 \pi^2}\zeta(2)+\frac{\gamma}{2 \pi^2}\eta(2)+\frac{\ln(2\pi)}{2 \pi^2}\zeta(2)+\frac{1}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\ln \left( k\right)}{k^2}+\frac{\ln \left( 2 \pi \right)}{2 \pi^2}\eta(2)-\frac{1}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\ln \left( k \right) \left(-1 \right)^{k}}{k^2}\\
&=\frac{ \ln\left(2 \pi\right)}{4} +\frac{\gamma}{12 }+\frac{\gamma}{24}+\frac{\ln(2\pi)}{12 }-\frac{1}{2 \pi^2}\zeta^{\prime}(2)+\frac{\ln \left( 2 \pi \right)}{24}-\frac{1}{2 \pi^2}\eta^{\prime}(2)\\
&=\frac{\gamma}{8}+\frac{3 \ln(2 \pi)}{8}-\frac{1}{2 \pi^2} \cdot \frac{\pi^2}{6}\left(\ln(2 \pi)+\gamma-12 \ln A \right)-\frac{1}{2 \pi^2} \left(\frac{\pi^2 \ln(2 )}{12} +\frac{\pi^2}{12}\left(\ln(2 \pi)+\gamma-12 \ln A\right)\right)\\
&=\frac{\gamma}{8}+\frac{3 \ln(2 \pi)}{8}-\frac{\ln(2 \pi)}{12}-\frac{\gamma}{12}+\ln A-\frac{\ln (2)}{24}-\frac{\ln(2 \pi)}{24}-\frac{\gamma}{24}+\frac{\ln A}{2}\\
&=\left(\frac{1}{8}-\frac{1}{12}-\frac{1}{24} \right)\gamma+\left(\frac38-\frac{1}{12}-\frac{1}{24} \right)\ln(2 \pi)-\frac{\ln(2)}{24}+\ln A+\frac{\ln A}{2}\\
&=\frac{3}{2}\ln A+ \frac{6}{24}\ln (2 \pi)-\frac{\ln (2)}{24}\\
&=\frac{3}{2}\ln A+\frac{5}{24}\ln 2+\frac{1}{4}\ln \pi \qquad \blacksquare
\end{aligned}$$
We used that
$\zeta^{\prime}(2)=\frac{\pi^2}{6}\left(\ln 2 \pi + \gamma -12 \ln A \right)$
And that
$\eta^{\prime}(s)=2^{1-s}\ln (2) \zeta(s)+(1-2^{1-s})\zeta^{\prime}(s)$
Best Answer
It should be
$$ \begin{aligned} \sum_{n=2}^{\infty}\frac{1}{(n^2-1)^2}&=\left(\frac{\psi^{\prime}(3)-\psi(3)}{4}+\frac{\psi^{\prime}(1)+\psi(1)}{4} \right)\\ &=\left(\frac{2\left(\frac{\pi^2}{6}-\color{red}{\frac54} \right)-2\left(-\gamma+\frac{3}{2} \right)}{8}+\frac{\frac{\pi^2}{3}-2\gamma}{8} \right)\\ &=\frac{\pi^2}{12}-\frac{11}{16} \end{aligned} $$