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It's clear that the integral is $\ds{\phi}$-independent.
${\phi \equiv {1 + \root{5} \over 2} > 0}$.
Namely,
$\ds{\int_{0}^{\infty}{1 - \expo{-\phi x} \over 1 + \expo{\phi x}}
\,{\dd x \over x}\ \stackrel{\phi x\ \to\ x}{=}\
\int_{0}^{\infty}{1 - \expo{-x} \over 1 + \expo{x}}\,{\dd x \over x}}$
\begin{align}
\int_{0}^{\infty}{1 - \expo{-x} \over 1 + \expo{x}}\,{\dd x \over x} & =
\int_{0}^{\infty}{1 - \expo{-x} \over 1 + \expo{-x}}\,\expo{-x}\,{\dd x \over x}
\ \stackrel{\expo{-x}\ =\ t}{=}\
-\int_{0}^{1}{1 - t \over 1 + t}\,{\dd t \over \ln\pars{t}}
\\[3mm] & =
\int_{0}^{1}{1 - t \over 1 + t}\
\overbrace{\int_{0}^{\infty}t^{\mu}\,\dd\mu}^{\ds{-\,{1 \over \ln\pars{t}}}}\
\,\dd t =
\int_{0}^{\infty}\int_{0}^{1}{t^{\mu} - t^{\mu + 1} \over 1 + t}\,\dd\mu\,\dd t
\\[3mm] & =
\int_{0}^{\infty}\bracks{%
2\int_{0}^{1}{t^{\mu} - t^{\mu + 1} \over 1 - t^{2}}\,\dd\mu -
\int_{0}^{1}{t^{\mu} - t^{\mu + 1} \over 1 - t}\,\dd\mu}
\\[3mm] & =
\int_{0}^{\infty}\bracks{%
\int_{0}^{1}{t^{\mu/2 - 1/2} - t^{\mu/2} \over 1 - t}\,\dd\mu -
\int_{0}^{1}{t^{\mu} - t^{\mu + 1} \over 1 - t}\,\dd\mu}
\\[8mm] & =
\int_{0}^{\infty}\left\lbrack%
\int_{0}^{1}{1 - t^{\mu/2} \over 1 - t}\,\dd\mu -
\int_{0}^{1}{1 - t^{\mu/2 - 1/2} \over 1 - t}\,\dd\mu\right.
\\[3mm] & \left.\mbox{} +
\int_{0}^{1}{1 - t^{\mu} \over 1 - t}\,\dd\mu -
\int_{0}^{1}{1 - t^{\mu + 1} \over 1 - t}\,\dd\mu\right\rbrack
\\[8mm] & =
\int_{0}^{\infty}\bracks{%
\Psi\pars{1 + {\mu \over 2}} - \Psi\pars{\half + {\mu \over 2}} +
\Psi\pars{1 + \mu} - \Psi\pars{2 + \mu}}\,\dd\mu
\\[3mm] & = \left.%
\ln\pars{\Gamma^{2}\pars{1 + \mu/2}\Gamma\pars{1 + \mu} \over
\Gamma^{2}\pars{1/2 + \mu/2}\Gamma\pars{2 + \mu}}
\right\vert_{\ 0}^{\ \infty}
\\[3mm] & = \underbrace{%
\lim_{\mu \to \infty}\ln\pars{\Gamma^{2}\pars{1 + \mu/2} \over
\Gamma^{2}\pars{1/2 + \mu/2}\pars{1 + \mu}}}_{\ds{-\ln\pars{2}}}\ -\
\underbrace{\ln\pars{\Gamma^{2}\pars{1}\Gamma\pars{1} \over
\Gamma^{2}\pars{1/2}\Gamma\pars{2}}}_{\ds{-\ln\pars{\pi}}}\ =\
\color{#f00}{\ln\pars{\pi \over 2}}
\end{align}
where $\Psi$ is the Digamma function and, by definition,
$\ds{\Psi\pars{z} = \totald{\ln\pars{\Gamma\pars{z}}}{z}}$. In the above calculation we used the well known identities ( $\gamma$ is the Euler-Mascheroni constant ):
\begin{align}
\Psi\pars{z} + \gamma & =
\int_{0}^{1}{1 - t^{z - 1} \over 1 - t}\,\dd t\,,\qquad\Re\pars{z} > 0
\\[3mm]
\Gamma\pars{1} & = \Gamma\pars{2} = 1\,,\quad\Gamma\pars{\half} = \root{\pi}
\,,\quad\Gamma\pars{z + 1} = z\,\Gamma\pars{z}
\end{align}
The last $\ds{\mu \to \infty}$ limit can be evaluated with Stirling Asymptotic Formula.
