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.
 A: Burniston and Siewert solved the equation:
$$ze^z=a(z+b)$$
Your equation can be reduced to the form:
$$5z-4= e^{z-1}z$$
which can be solved by Burniston Siewert integral
== References ==
[68] C. E. Siewert and E. E. Burniston, "Solutions of the Equation $ze^z=a(z+b)$," Journal of Mathematical Analysis and Applications, 46 (1974) 329-337. 
http://www4.ncsu.edu/~ces/pdfversions/68.pdf
A: As said in comments, there is no closed form for the solution.
However, we can generate closed forms of very good approximations starting with as estimate $x_0$ written as a rational number and apply one single iteration of Newton method to get $x_1$ given, for the case of this function, by $$x_1=\frac{(16-9 x_0) x_0-(x_0-4)^2 (x_0+1) \log (x_0+1)}{(x_0-12) x_0+12}$$ Using $x_0=\frac {11}5$ as you gave in the post this will lead to $$x_1=\frac{1045+1296 \log \left(\frac{16}{5}\right)}{1195}\approx 2.135935941$$ Using this result, make $x_0=\frac {2136}{1000}$ which leads to $$x_1=\frac{13450125+21281288 \log \left(\frac{392}{125}\right)}{17713875}\approx 2.132426289$$ and you could repeat this process as long as you wish.
A: 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 $\frac {4z} {z-1}$
