Is there any integral for the Golden Ratio?

Potentially interesting:

$$\log\varphi=\int_0^{1/2}\frac{dx}{\sqrt{x^2+1}}$$

Perhaps also worthy of consideration:

$$\arctan \frac{1}{\varphi}=\frac{\int_0^2\frac{1}{1+x^2}\, dx}{\int_0^2 dx}=\frac{\int_{-2}^2\frac{1}{1+x^2}\, dx}{\int_{-2}^2 dx}$$

A development of the first integral:

$$\log\varphi=\frac{1}{2n-1}\int_0^{\frac{F_{2n}+F_{2n-2}}{2}}\frac{dx}{\sqrt{x^2+1}}$$

$$\log\varphi=\frac{1}{2n}\int_1^{\frac{F_{2n+1}+F_{2n-1}}{2}}\frac{dx}{\sqrt{x^2-1}}$$

which stem from the relationship $(x-\varphi^m)(x-\bar\varphi^m)=x^2-(F_{m-1}+F_{m+1})x+(-1)^m$, where $\bar\varphi=\frac{-1}{\varphi}=1-\varphi$ and $F_k$ is the $k$th Fibonacci number. I particularly enjoy:

$$\log\varphi=\frac{1}{3}\int_0^{2}\frac{dx}{\sqrt{x^2+1}}$$ $$\log\varphi=\frac{1}{6}\int_1^{9}\frac{dx}{\sqrt{x^2-1}}$$ $$\log\varphi=\frac{1}{9}\int_0^{38}\frac{dx}{\sqrt{x^2+1}}$$ $$\log\varphi=\frac{1}{12}\int_1^{161}\frac{dx}{\sqrt{x^2-1}}$$

In this answer, it is shown that $$ \int_0^\infty\frac{\sqrt{x}}{x^2+2x+5}\mathrm{d}x=\frac\pi{2\sqrt\phi} $$

[EDIT:20190112 Another connection between $\pi$ and $\phi$ at the bottom] An identity derived from the Rogers-Ramanujan continued fraction ($R(q)$, not defined here) exhibits a $\phi$ factor:

$$ \frac{1}{(\sqrt{\phi\sqrt{5}})e^{2\pi/5}} = 1+\frac{e^{-2\pi}}{1+\frac{e^{-4\pi}}{1+\frac{e^{-6\pi}}{1+\frac{e^{-8\pi}}{1+\frac{e^{-10\pi}}{1+\frac{e^{-12\pi}}{\cdots}}}}}}$$

and one can then obtain a formula like: $$ \ln \left( \sqrt{4\phi+3}-\phi^2\right) = -\frac{1}{5}\int_{e^{-2\pi}}^1 \frac{(1-t)^5(1-t^2)^5(1-t^3)^5 \dots}{(1-t^5)(1-t^{10})(1-t^{15}) \dots}\frac{dt}{t}$$ which beautifully links integrals, $e$, $\phi$ and $\pi$. It is described for instance in Golden Ratio and a Ramanujan-Type Integral.

Not very practical though to obtain $\phi$ rational approximations.

[EDIT] In M. D. Hirschhorn, A connection between $\pi$ and $\phi$, Fibonacci Quarterly, 2015, another asymptotic relation is:

$$ \frac{1}{\pi}=\lim_{n\to \infty} 2n {5}^{1/4}\sum_{k=0}^{n}\binom{n}{k}^2\binom{n+k}{k}/\phi^{5n+5/2}$$