Evaluate $\int_0^{\infty}\frac{e^x-1}{xe^x(e^x+1)}dx.$ 
Evaluate the integral
  $$
\int_{0}^{\infty}{\mathrm{e}^{x} - 1 \over x\,\mathrm{e}^{x}\left(\mathrm{e}^x+1\right)}\,\mathrm{d}x\,.
$$

I have no idea how to approach integrals like this and I can't get any valuable result. Any hints will be appreciated.

Notice that
\begin{align*}
\int_{0}^{\infty} \frac{e^x - 1}{x e^x(e^x + 1)} \, dx
&= \int_{0}^{\infty} \left( \int_{0}^{1} e^{xt} \, dt \right) \frac{dx}{e^x(e^x+1)} \\
&= \int_{0}^{1} \int_{0}^{\infty} \frac{e^{tx}}{e^x(e^x+1)} \, dxdt.
\end{align*}
In order to compute the inner integral, we utilize the geometric series as follows:
\begin{align*}
\int_{0}^{\infty} \frac{e^{tx}}{e^x(e^x+1)} \, dx
&= \int_{0}^{\infty} \frac{e^{-(2-t)x}}{1+e^{-x}} \, dx \\
&= \sum_{n=0}^{\infty} (-1)^n \int_{0}^{\infty} e^{-(n+2-t)x} \, dx \\
&= \sum_{n=0}^{\infty}\frac{(-1)^n}{n+2-t}. \tag{1}
\end{align*}
Plugging this back, we have
\begin{align*}
\int_{0}^{\infty} \frac{e^x - 1}{x e^x(e^x + 1)} \, dx
&= \sum_{n=0}^{\infty} (-1)^n\int_{0}^{1} \frac{dt}{n+2-t} \\
&= \sum_{n=0}^{\infty} (-1)^n ( \log(n+2) - \log(n+1) ) \\
&= \log \left( \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdots \right) \\
&= \log \left(\frac{\pi}{2}\right),
\end{align*}
where in the last line we utilized the Wallis product. Alternatively, we can identify $\text{(1)}$ as
$$ \int_{0}^{\infty} \frac{e^{tx}}{e^x(e^x+1)} \, dx = \frac{1}{2}\left( \psi_0\left(\frac{3-t}{2}\right) - \psi_0\left(\frac{2-t}{2}\right) \right), $$
where $\psi_0$ is the digamma function. Plugging this back gives
$$ \int_{0}^{\infty} \frac{e^x - 1}{x e^x(e^x + 1)} \, dx = \log\left( \frac{\Gamma\left(\frac{3}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma(1)^2} \right) = \log\left(\frac{\pi}{2}\right). $$
