Integrals of the form $\int_0^1\left(\frac{1}{x+1}-\ln\left|\frac{x+1}{2}\right|-\ln\ln\left|\frac{x+1}{x-1}\right|\right)dx$ 
How could we evaluate the following integrals? (Integrands pictures above)
Blue:
\begin{align}
\int_0^1 \left(\frac{1}{x+1}-\ln\left|\frac{x+1}{2}\right|-\ln\ln\left|\frac{x+1}{x-1}\right|\right)dx &= \gamma - \ln\pi + \ln 2 + 1 \\
\int_0^\infty \left(\frac{1}{x+1}-\ln\left|\frac{x+1}{2}\right|-\ln\ln\left|\frac{x+1}{x-1}\right|\right)dx &= -1 + \ln 2
\end{align}
Red:
$$\begin{align}
\int_0^1 \left(\frac{1}{x+1}+\ln\left|\frac{x-1}{2}\right|+\ln\ln\left|\frac{x+1}{x-1}\right|\right)dx &= -\gamma + \ln\pi - \ln 2 - 1 \\
\int_0^\infty \left(\frac{1}{x+1}+\ln\left|\frac{x-1}{2}\right|+\ln\ln\left|\frac{x+1}{x-1}\right|\right)dx &= -1 - \ln 2
\end{align}$$
 A: A solution using Riemann $\zeta$, with $\zeta(0)=-1/2$ and $\zeta'(0)/\zeta(0)=\ln2\pi$. Let \begin{align}
F(a,b)&=\int_a^b\left(\frac1{x+1}+\ln\left|\frac{x-1}{2}\right|+\ln\ln\left|\frac{x+1}{x-1}\right|\right)dx,\\
G(a,b)&=\int_a^b\left(\frac1{x+1}-\ln\left|\frac{x+1}{2}\right|-\ln\ln\left|\frac{x+1}{x-1}\right|\right)dx,\\
S(a,b)&=F(a,b)+G(a,b)=\int_a^b\left(\frac2{x+1}-\ln\left|\frac{x+1}{x-1}\right|\right)dx,\\
D(a,b)&=F(a,b)-G(a,b)=\int_a^b\left(\ln\left|\frac{x^2-1}4\right|+2\ln\ln\left|\frac{x+1}{x-1}\right|\right)dx.
\end{align}
Here $S(a,b)$ is elementary, and we easily find $S(0,1)=0$ and $S(1,\infty)=-2$.
Similarly, after elementary integration, we get $D(0,1)=2I-2$, where $$I=\int_0^1\ln\ln\frac{1+x}{1-x}\,dx=2\int_0^\infty\frac{e^t\ln t\,dt}{(e^t+1)^2}$$ after $x=(e^t-1)/(e^t+1)$. Thus $I=2f'(0)=f'(0)/f(0)$ where, for $\Re s>-1$, $$f(s)=\int_0^\infty\frac{t^s e^t\,dt}{(e^t+1)^2}=\Gamma(s+1)\eta(s)$$ with $\eta(s)=(1-2^{1-s})\zeta(s)$; this can be obtained for $\Re s>0$ using integration by parts, and then for $\Re s>-1$ by analytic continuation. With $\zeta(0)$ and $\zeta'(0)$ at hand, we obtain $I=\ln(\pi/2)-\gamma$.
For $D(1,\infty)$, we substitute $x=(e^t+1)/(e^t-1)$ and integrate by parts: \begin{align}
D(1,\infty)&=2\int_0^\infty\ln\frac{t^2 e^t}{(e^t-1)^2}\,d\frac{-1}{e^t-1}=2g(0),\\g(s)&=\int_0^\infty\left(1+\frac2t-\frac{2e^t}{e^t-1}\right)\frac{t^s\,dt}{e^t-1}.
\end{align}
This time we get $g(s)=\Gamma(s)\big(s\zeta(s+1)+2(1-s)\zeta(s)\big)$, for $\Re s>1$ via integration by parts, then for $\Re s>-1$ by analytic continuation, and finally $g(0)=\gamma+1-\ln2\pi$.
Now we have all the pieces to gather the results from.
