Evaluation of $\sum_{n=1}^\infty \frac{(-1)^{n-1}\eta(n)}{n} $ without using the Wallis Product In THIS ANSWER, I showed that 
$$2\sum_{s=1}^{\infty}\frac{1-\beta(2s+1)}{2s+1}=\ln\left(\frac{\pi}{2}\right)-2+\frac{\pi}{2}$$
where $\beta(s)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s}$ is the Dirichlet Beta Function.
In the development, it was noted that
$$\begin{align}
\sum_{n=1}^\infty(-1)^{n-1}\log\left(\frac{n+1}{n}\right)&=\log\left(\frac21\cdot \frac23\cdot \frac43\cdot \frac45\cdots\right)\\\\
&=\log\left(\prod_{n=1}^\infty \frac{2n}{2n-1}\frac{2n}{2n+1}\right)\\\\
&=\log\left(\frac{\pi}{2}\right) \tag 1
\end{align}$$
where I used Wallis's Product for $\pi/2$.

If instead of that approach, I had used the Taylor series for the logarithm function, then the analysis would have led to 
$$\sum_{n=1}^\infty(-1)^{n-1}\log\left(\frac{n+1}{n}\right)=\sum_{n=1}^\infty \frac{(-1)^{n-1}\eta(n)}{n} \tag 2$$
where $\eta(s)=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}$ is the Dirichlet eta function.
Given the series on the right-hand side of $(2)$ as a starting point, it is evident that we could simply reverse steps and arrive at $(1)$.  

But, what are some other distinct ways that one can take to evaluate the right-hand side of $(2)$? 

For example, one might try to use the integral representation
$$\eta(s)=\frac{1}{\Gamma(s)}\int_0^\infty \frac{x^{s-1}}{1+e^x}\,dx$$
and arrive at 
$$\sum_{n=1}^\infty \frac{(-1)^{n-1}\eta(n)}{n} =\int_0^\infty \frac{1-e^{-x}}{x(1+e^x)}\,dx =\int_1^\infty \frac{x-1}{x^2(x+1)\log(x)}\,dx \tag 3$$
Yet, neither of these integrals is trivial to evaluate (without reversing the preceding steps).

And what are some other ways to handle the integrals in $(3)$?

 A: 
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
$$
\int_0^\infty \frac{1-e^{-x}}{x(e^x+1)}dx=\log \left(\frac{\pi}2\right).\tag5
$$
A: 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 $$\sum_{n\geq1}\frac{\left(-1\right)^{n-1}}{n}\eta\left(n\right)=\log\left(2\right)+2\log\left(\frac{\sqrt{\pi}}{2}\right)=\log\left(\frac{\pi}{2}\right).$$
A: 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.
