NEW ANSWER. I found yet another way of solving this problem. My new solution does not use contour integration, and is based on the following observation: for $|z| \leq 1$,
$$ - \int_{-1}^{1} \frac{1}{x} \sqrt{\frac{1+x}{1-x}} \log(1 - zx) \, dz= \pi \sin^{-1} z - \pi \log \left( \tfrac{1}{2}+\tfrac{1}{2}\sqrt{1-z^{2}} \right) . $$
As I want to keep both the old answer and the new answer, I posted my new solution to other page. You can check it here.
OLD ANSWER. Okay here is another solution. It is also related to my generalization.
We claim the following proposition:
Proposition. If $0 < r < 1$ and $r < s$, then
$$ I(r, s) := \int_{-1}^{1} \frac{1}{x} \sqrt{\frac{1+x}{1-x}} \log \left( \frac{1 + 2rsx + (r^{2} + s^{2} - 1)x^{2}}{1 - 2rsx + (r^{2} + s^{2} - 1)x^{2}} \right) \, dx = 4\pi \arcsin r. \tag{1} $$
Assuming this proposition, all that we have to do is to solve the non-linear system of equations
$$ 2rs = 2 \quad \text{and} \quad r^{2} + s^{2} - 1 = 2. $$
The unique solution satisfying the condition of the proposition is $r = \phi - 1$ and $s = \phi$. So by $\text{(1)}$ we have
\begin{align*}
\int_{-1}^{1} \frac{1}{x} \sqrt{\frac{1+x}{1-x}} \log \left( \frac{1 + 2x + 2x^{2}}{1 - 2x + 2x^{2}} \right) \, dx
& = I(\phi-1, \phi) \\
&= 4\pi \arcsin (\phi - 1)
= 4\pi \operatorname{arccot} \sqrt{\phi}.
\end{align*}
Thus it remains to prove the proposition.
Proof of Proposition. We divide the proof into several steps.
Step 1. (Case reduction by analytic continuation) We first remark that given $r$ and $s$, we always have
$$ \min_{-1 \leq x \leq 1} \{ 1 \pm 2rsx + (r^{2} + s^{2} - 1)x^{2} \} > 0. \tag{2} $$
Indeed, it is not hard to check if we utilize the following equality
$$ 1 \pm 2rsx + (r^{2} + s^{2} - 1)x^{2} = (1 \pm rsx)^{2} - (1 - r^{2})(1 - s^{2}) x^{2}. $$
Then $\text{(2)}$ shows that the integrand of $I(r, s)$ remains holomoprhic under small perturbation of $s$ in $\Bbb{C}$. So it allows us to extend $s \mapsto I(r, s)$ as a holomorphic function on some open set containing the line segment $(r, \infty) \subset \Bbb{C}$. Then by the principle of analytic continuation, it is sufficient to prove that $\text{(1)}$ holds for $r < s < 1$.
Step 2. (Integral representation of $I$) Put $r = \sin \alpha$ and $s = \sin \beta$, where $ 0 < \alpha < \beta < \frac{\pi}{2}$. Then
\begin{align*}
I(r, s)
&= \int_{-1}^{1} \frac{1+x}{x\sqrt{1-x^{2}}} \log \left( \frac{1 + 2rsx + (r^{2} + s^{2} - 1)x^{2}}{1 - 2rsx + (r^{2} + s^{2} - 1)x^{2}} \right) \, dx \\
&= \int_{0}^{1} \frac{2}{x\sqrt{1-x^{2}}} \log \left( \frac{1 + 2rsx + (r^{2} + s^{2} - 1)x^{2}}{1 - 2rsx + (r^{2} + s^{2} - 1)x^{2}} \right) \, dx \qquad (\because \text{ parity}) \\
&= \int_{1}^{\infty} \frac{2}{\sqrt{x^{2}-1}} \log \left( \frac{x^{2} + 2rsx + (r^{2} + s^{2} - 1)}{x^{2} - 2rsx + (r^{2} + s^{2} - 1)} \right) \, dx \qquad (x \mapsto x^{-1}) \\
&= \int_{0}^{1} \frac{2}{t} \log \left( \frac{\left(t+t^{-1}\right)^{2} + 4rs\left(t+t^{-1}\right) + 4(r^{2} + s^{2} - 1)}{\left(t+t^{-1}\right)^{2} - 4rs\left(t+t^{-1}\right) + 4(r^{2} + s^{2} - 1)} \right) \, dt,
\end{align*}
where in the last line we utilized the substitution $x = \frac{1}{2}(t + t^{-1})$. If we introduce the quartic polynomial
\begin{align*}
p(t) = t^{4} + 4rst^{3} + (4r^{2}+4s^{2}-2)t^{2} + 4rst + 1,
\end{align*}
then by the property $p(1/t) = t^{-4}p(t)$, we can simplify
\begin{align*}
I(r, s)
&= 2 \int_{0}^{1} \frac{\log p(t) - \log p(-t)}{t} \, dt
= \int_{0}^{\infty} \frac{\log p(t) - \log p(-t)}{t} \, dt \\
&= - \int_{0}^{\infty} \left( \frac{p'(t)}{p(t)} + \frac{p'(-t)}{p(-t)} \right) \log t \, dt
= - \frac{1}{2} \Re \int_{-\infty}^{\infty} \left( \frac{p'(z)}{p(z)} + \frac{p'(-z)}{p(-z)} \right) \log z \, dz,
\end{align*}
where we choose the branch cut of $\log$ in such a way that it avoids the upper-half plane
$$\Bbb{H} = \{ z \in \Bbb{C} : \Im z > 0 \}.$$
Step 3. (Residue calculation) Since
$$ f(z) := \left( \frac{p'(z)}{p(z)} + \frac{p'(-z)}{p(-z)} \right) \log z = O\left(\frac{\log z}{z^{2}} \right) \quad \text{as } z \to \infty, $$
by replacing the contour of integration by a semicircle of sufficiently large radius, it follows that
\begin{align*}
I(r, s) = - \frac{1}{2} \Re \left\{ 2 \pi i \sum_{z_{0} \in \Bbb{H}} \operatorname{Res}_{z = z_{0}} f(z) \right\} = \pi \Im \sum_{z_{0} \in \Bbb{H}} \operatorname{Res}_{z = z_{0}} f(z).
\end{align*}
(It turns out that $f(z)$ has only logarithmic singularity at the origin. So it does not account for the value of $I(r, s)$.) But by a simple calculation, together with the condition $ 0 < \alpha < \beta < \frac{\pi}{2}$, we easily notice that the zeros of $p(z)$ are exactly
$$ e^{\pm i(\alpha + \beta)} \quad \text{and} \quad -e^{\pm i(\alpha - \beta)}. $$
Now let $Z_{+}$ be the set of zeros of $p(z)$ in $\Bbb{H}$ and $Z_{-}$ be the set of zeros of $p(z)$ in $-\Bbb{H}$. Then
$$ Z_{+} = \{ e^{i(\beta+\alpha)}, -e^{-i(\beta - \alpha)} \} \quad \text{and} \quad Z_{-} = \{ e^{-i(\beta+\alpha)}, -e^{i(\beta- \alpha)} \}. $$
This in particular shows that
$$ \frac{p'(z)}{p(z)}\log z = \sum_{z_{0} \in Z_{+}} \frac{\log z}{z - z_{0}} + \text{holomorphic function on } \Bbb{H} $$
and
$$ \frac{p'(-z)}{p(-z)}\log z = -\sum_{z_{0} \in -Z_{-}} \frac{\log z}{z - z_{0}} + \text{holomorphic function on } \Bbb{H}. $$
So it follows that
\begin{align*}
I(r, s)
&= \pi \Im \left\{ \sum_{z_{0} \in Z_{+}} \log z_{0} - \sum_{z_{0} \in -Z_{-}} \log z_{0} \right\} \\
&= \pi \Im \left\{ \log e^{i(\beta+\alpha)} + \log e^{i(\pi-\beta+\alpha)} - \log e^{i(\pi-\beta-\alpha)} - \log e^{i(\beta-\alpha)} \right\} \\
&= \pi \Im \left\{ i(\beta+\alpha) + i(\pi-\beta+\alpha) - i(\pi-\beta-\alpha) - i(\beta-\alpha) \right\} \\
&= 4\pi \alpha
= 4\pi \arcsin r.
\end{align*}
This completes the proof.
