My original problem is $$g(r)=\ln(1+r)-a\sqrt{1-\frac{1}{(1+r)^2}}$$ where $r\geq0$ is a variable, $a\geq0$ is a constant. And I'm trying to find the root where $g(r)=0$. This is an transcendental equation, maybe it's hard to find the analytic solution, so I'm trying to find the approximate analytic solution.
What I have tried:
Letting $x=\ln(1+r)\geq0$, we can obtain$$x-a\sqrt{1-e^{-2x}}=0$$ $$x^2=a^2(1-e^{-2x})$$ $$e^{-2x}=1-\frac{1}{a^2}x^2=\frac{1}{a^2}(x^2-a^2)$$ then, letting $f(x)=e^{-2x}+\frac{1}{a^{2}}(x+a)(x-a)$, where $a\geq0$, $x\geq0$.
I have tried to find the root using Lambert W function, but I don't know how to transform the equation to the form $xe^x=c$.
Can I find the approximate analytic solution where $f(x)=0$ using Lambert W function? Or is there any other way to solve $g(r)=0$?
Now we can obtain better approximate analytic solution when $a\geq2$, but the approximation error of $g(r)$ is big than $10^{−2}$ for $2\geq a \geq0$. How to further obtain a better approximate analytic solution for $2\geq a \geq0$?
Consider instead that you look for the zero of function $$f(r)=\frac{(r+1) \log (r+1)}{\sqrt{r (r+2)}}-a$$
If $a$ is large, make a series expansion $$\frac{(r+1) \log (r+1)}{\sqrt{r (r+2)}}=\log (r)+\frac{1}{r}+\frac{\log (r)-1}{2 r^2}+\frac{5-6\log (r)}{6r^3}+O\left(\frac{1}{r^4}\right)$$ So, a first approximation is given by $$\log (r)+\frac{1}{r}=a \implies r_0=-\frac{1}{W\left(-e^{-a}\right)}$$ and here is Lambert function.
Using then next term of the expansion would give $$\log(r_1)=\frac{2 a r_0^2-2 r_0+1}{2 r_0^2+1}\implies r_1=\exp\left(\frac{2 a r_0^2-2 r_0+1}{2 r_0^2+1} \right)$$
A few numbers for illustration
$$\left( \begin{array}{cccc} a & r_0 & r_1 & \text{exact} \\ 1.00 & 1.00000 & 1.39561 & 1.50068 \\ 1.25 & 2.22825 & 2.26915 & 2.28989 \\ 1.50 & 3.31445 & 3.28596 & 3.30121 \\ 1.75 & 4.63863 & 4.58267 & 4.59469 \\ 2.00 & 6.30540 & 6.23985 & 6.24911 \\ 2.25 & 8.42598 & 8.35958 & 8.36650 \\ 2.50 & 11.1362 & 11.0733 & 11.0784 \\ 2.75 & 14.6076 & 14.5503 & 14.5539 \\ 3.00 & 19.0588 & 19.0079 & 19.0104 \\ 3.25 & 24.7699 & 24.7254 & 24.7271 \\ 3.50 & 32.0997 & 32.0613 & 32.0625 \\ 3.75 & 41.5089 & 41.4761 & 41.4769 \\ 4.00 & 53.5888 & 53.5610 & 53.5615 \end{array} \right)$$
Edit
If you notice that $r(r+2) \sim (r+1)^2$, you have the easy estimate $r_0=e^a-1$ and the first iterate of Newton method is $$r_1=e^a-1-\frac{a \left(e^{2 a}-1\right) \left(e^a-\sqrt{e^{2 a}-1}\right)}{e^{2 a}-a-1}$$ which is quite accurate.
For example, $a=1$ gives $r_1=1.49229$ (the exact solution being $1.50068$) and $a=2$ gives $r_1=6.24783$ (the exact solution being $6.24911$