Exact values of the equation $\ln (x+1)=\frac{x}{4-x}$ I'm asking for a closed form (an exact value) of the equation solved for $x$
$$\ln (x+1)=\frac{x}{4-x}$$
$0$ is trivial but there is another solution (approximately 2,2...).
I've tried with Lambert's W Function or with Regularized Gamma Functions but I didn't get so far.

I didn't sleep well since the day this problem came into my life so, working really hard, I personally found how to solve this kind of problems with a generalization of Lambert's function, here's my own answer:
$\ln(x+1)={x\over {4-x}}$
let $y=x+1$
$\ln(y)=-{{1-y}\over {5-y}}$
$y=e^{-{{1-y}\over {5-y}}}$
let $z={{1-y}\over {5-y}}$
$e^{-z}={{5z-1}\over {z-1}}$
$\frac 1 5={{z-\frac 1 5}\over {z-1}}e^z$
which is an equation of the form $e^{cx}{{x-a}\over {x-b}}=k$ that has solution(s):
$x=a+\frac 1 cW_{-ke^{-ca}}(cke^{-ca}(a-b))$
so we can now find that $z$ is equal to:
$z=\frac 1 5+W_{-\frac 1 5 e^{-\frac 1 5}}(-\frac 4 {25}e^{-\frac 1 5})$
So the solutions of the equation for $x$ are $0$ and $\frac {4z} {z-1}$
