Evaluation of $\sum_{n=1}^\infty \frac{(-1)^{n-1}\eta(n)}{n} $ without using the Wallis Product
What are some other ways to handle the following integral? $$ \int_0^\infty \frac{1-e^{-x}}{x(1+e^x)}\,dx \tag 1 $$
One may set $$ I(s):=\int_0^\infty \frac{1-e^{-sx}}{x(e^x+1)}dx, \quad s>0. \tag2 $$ We may differentiate under the integral sign, in order to get rid of the factor $x$ in the denominator, obtaining $$ \begin{align} I'(s)&=\int_0^\infty \frac{e^{-sx}}{e^x+1}dx \\\\I'(s)&=\int_0^\infty e^{-(s+1)x}\sum_{n=0}^\infty(-1)^n e^{-nx} dx \\\\I'(s)&=\sum_{n=0}^\infty(-1)^n\int_0^\infty e^{-(n+s+1)x} dx \\\\I'(s)&=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n+s} \\\\I'(s)&=\frac12\psi\left(1+\frac{s}2\right)-\frac12\psi\left(\frac{1+s}2\right) \tag3 \end{align} $$ where $\displaystyle \psi : = \Gamma'/\Gamma$.
Integrating $(3)$, with the fact that, as $s \to 0$, $I(s) \to 0$, we get
$$ \int_0^\infty \frac{1-e^{-sx}}{x(e^x+1)}dx=\frac12\log(\pi)+\log \Gamma\left(1+\frac{s}2\right)-\log \Gamma\left(\frac{1+s}2\right), \quad s>0. \tag4 $$
By putting $s:=1$ in $(4)$, one obtains $$ \int_0^\infty \frac{1-e^{-x}}{x(e^x+1)}dx=\log \left(\frac{\pi}2\right).\tag5 $$
Another way to handle $(2)$ is using the identity $$\eta\left(s\right)=\left(1-\frac{1}{2^{s-1}}\right)\zeta\left(s\right) $$ hence, since $\eta\left(1\right)=\log\left(2\right) $, $$\sum_{n\geq1}\frac{\left(-1\right)^{n-1}}{n}\eta\left(n\right)=\log\left(2\right)+\sum_{n\geq2}\frac{\left(-1\right)^{n-1}}{n}\eta\left(n\right) $$ $$=\log\left(2\right)+\sum_{n\geq2}\frac{\left(-1\right)^{n-1}}{n}\zeta\left(n\right)-\sum_{n\geq2}\frac{\zeta\left(n\right)}{n}\left(-\frac{1}{2}\right)^{n-1} $$ and now we can use the identity $$\sum_{n\geq2}\frac{\zeta\left(n\right)}{n}\left(-x\right)^{n}=x\gamma+\log\left(\Gamma\left(x+1\right)\right),\,-1<x\leq1 $$ which can be proved taking the log of the Weierstrass product of Gamma. So $$\sum_{n\geq1}\frac{\left(-1\right)^{n-1}}{n}\eta\left(n\right)=\log\left(2\right)+2\log\left(\frac{\sqrt{\pi}}{2}\right)=\log\left(\frac{\pi}{2}\right).$$
Observation $1$:
A suggestion made in a comment from @nospoon was to expand one of the integrals in a series and exploit Frullani's Integral. Proceeding accordingly, we find that
$$\begin{align} \int_0^\infty \frac{1-e^{-x}}{x(1+e^x)}\,dx&=\int_0^\infty \left(\frac{(e^{-x}-e^{-2x})}{x}\right)\left(\sum_{n=0}^\infty (-1)^{n}e^{-nx}\right)\,dx\\\\ &=\sum_{n=0}^\infty (-1)^{n} \int_0^\infty \frac{e^{-(n+1)x}-e^{-(n+2)x}}{x}\,dx\\\\ &=\sum_{n=0}^\infty (-1)^{n} \log\left(1+\frac{1}{n+1}\right)\\\\ &=\sum_{n=1}^\infty (-1)^{n-1} \log\left(1+\frac{1}{n}\right) \end{align}$$
thereby recovering the left-hand side of Equation $(1)$ in the OP.
Observation $2$:
In the answer posted by Olivier Oloa, note the intermediate relationship
$$I'(s)=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n+s}$$
Upon integrating $I'(s)$, as $\int_0^1 I'(s)\,ds$, we find that
$$I(1)=\sum_{n=1}^\infty (-1)^{n-1}\log\left(1+\frac1n\right)$$
thereby recovering again the left-hand side of Equation $(1)$ in the OP.