How to find a solution to the following equation \begin{align*} \frac{1}{1+x}-\frac{c}{x}-2\log \left( \frac{1+x}{x}\right)+A=0 \end{align*}
where $c$ and $A$ are some constants such that $c\ge 1$ and $A>0$.
I tried a few softwares to solve it but I don't think there is an analytic solution.
Could some one please suggest a way to find solution in some approximate sense? Thank you
Rearranging, probably useless but I'll leave it here:
$$2\log \left(\frac{1+x}{x}\right) = \frac{x}{x^2+x} - \frac{cx + c}{x^2+x} + \frac{Ax^2 + Ax}{x^2 + x} $$
$$\log \left(\frac{1+x}{x}\right) = \frac{Ax^2 + (A+1-c)x - c}{2x^2 + 2x}$$
$$\frac{1+x}{x} = \exp\left(\frac{Ax^2 + (A+1-c)x - c}{2x^2 + 2x}\right)$$