Functional "Square Root" of $1/(x+1)$

Since $h(x)=\frac{1}{x+1}$ is a fractional linear map, it is enough to devise a matrix square root for $\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}$ to have a functional square root for $h(x)$.

If I am not wrong, this approach gives $$ \frac{\sqrt{1-2 i}+x\sqrt{-1+2 i} }{\sqrt{4+2 i}+x\sqrt{1-2 i}} = \color{red}{\frac{1+ix}{(1+i)+x}}$$ as a functional square root of $\frac{1}{1+x}$.

With an explorative approach we could first look at rational expressions of the iterates to positive integer heights and try, whether this gives a scheme for a continuous interpolation.
What we get for the first few iterates is $$ f°^1(x) = {1+0x\over 1+1x} \qquad f°^2(x) = {1+1x\over 2+1x} \qquad f°^3(x) = {2+1x\over 3+2x} \qquad f°^4(x) = {3+2x\over 5+3x} \cdots $$ and a pattern of Fibonacci-numbers $fib_k$ is much suggestive (and can also be proven) $$ f°^h(x) = {fib_h+fib_{h-1}x\over fib_{h+1} + fib_h x} \tag 1 $$ Since the Fibonacci-numbers have a generalization to continuous indexes we have the option to define the fractional iterates based on this generalized Fibonacci-numbers.

---

The most common$\,^{[1]}$ Fibonacci-generalization is that based on the Binet-formula: Let $\Phi = {1+\sqrt 5 \over 2} \approx 1.618... $ then the Binet-formula gives $$ fib_h = {\Phi^h - (1-\Phi)^h\over \sqrt 5} \tag 2$$ If $h=0.5$ as asked here, we have to use $fib_{-0.5}, fib_{0.5},fib_{1.5} $ in formula (1). That three generalized Fibonacci-values are to be inserted $$ fib_{-0.5} \approx 0.351577584254 + 0.568864481006 î \\ fib_{ 0.5} \approx 0.568864481006 - 0.351577584254 î \\ fib_{ 1.5} \approx 0.920442065260 + 0.217286896752 î \\ $$ A massive reduction of the explicite formula (1) when that values are inserted, however, can be done when we simply cancel $fib_h$. We have then formally $$ f°^h(x) = {1 +{fib_{h-1}\over fib_h}x\over {fib_{h+1} \over fib_h}+ 1x} \tag 3 $$ and for $h=0.5$ we get here the handy coefficients ${fib_{h-1}\over fib_h} = 1 î $ and ${fib_{h+1}\over fib_h} = 1 + 1 î $ such that we can write the compacted formula $$ f°^{0.5}(x) = {1 + 1 î x\over (1 + 1 î )+ 1x} \tag 4 $$ which is identical to that already given solution of Jack D'Aurizio.

---

The intention of this answer is to show one way, how to generalize that fractional iteration formula to other values than actually $h=1/2$ only.

_{$\,^{[1]}$ There is also a real-to-real-interpolation common (see wikipedia; in an earlier thread here in MSE I've discussed the difference between that two methods); I didn't check what we would get here with that method}