<s>[INST] $I_1=\int_0^1 \frac 1 {1+\frac 1 {\sqrt x}} dx$ If we let $I_2= \int_0^1 \frac 1 {1+\frac 1 {1+\sqrt x}} dx$ and so on, we’re tasked with solving $I_n$.  We can show that $I_2 = \int \frac {1+\sqrt x} {1-\sqrt x + 1} = \int \frac {1+\sqrt x} {2+\sqrt x }$
And that $I_3=\int \frac {2+\sqrt x}{3+2\sqrt x}$, and that in general, using $F_n$ to denote the nth Fibonacci number, 
$$I_n=\int^1_0 \frac{F_n+F_{n-1}\sqrt x}{F_{n+1}+F_n \sqrt x}dx$$
However, at this point I’m uncertain how to proceed; we obviously have some ratio of Fibonacci’s which looks like the golden ratio, but I’m unsure what substitutions to make to have the golden ratio more clearly emerge as a useful quantity.
 [/INST] We define $$j=j(a,b)=\int_0^1\frac{a+\sqrt x}{b+\sqrt x}dx$$
So we see that 
$$j^*(a_1,a_2,a_3,a_4)=\int_0^1\frac{a_1+a_2\sqrt x}{a_3+a_4\sqrt x}dx=\frac{a_2}{a_4}j\left(\frac{a_1}{a_2},\frac{a_3}{a_4}\right)$$
So then we see that 
$$\begin{align}
j&=\int_0^1\frac{a-b+b+\sqrt x}{b+\sqrt x}dx\\&=(a-b)\int_0^1\frac{dx}{b+\sqrt x}+\int_0^1\frac{b+\sqrt x}{b+\sqrt x}dx\\&=(a-b)\int_0^1\frac{dx}{b+\sqrt x}+1
\end{align}$$
Then we set $x=u^2\Rightarrow dx=2udu$:
$$\begin{align}
\int_0^1\frac{dx}{b+\sqrt x}&=2\int_0^1\frac{u}{b+u}du\\
&=2\int_0^1\frac{-b+b+u}{b+u}du\\
&=2-2b\int_0^1\frac{du}{b+u}du\\
&=2-2b\ln|u+b|]_0^1\\
&=2+2b\ln\left|\frac{b}{b+1}\right|
\end{align}$$
So $$j(a,b)=1+2(a-b)\left(1+b\ln\left|\frac{b}{b+1}\right|\right)$$
And your integral is given by $$I_n=j^*(F_n,F_{n-1},F_{n+1},F_n)=\frac{F_{n-1}}{F_n}j\left(\frac{F_n}{F_{n-1}},\frac{F_{n+1}}{F_n}\right)$$
</s>