A closed form for: $\int_{0}^{\infty} \frac{1}{(x-\log x)^2}dx$ Is it possible to find a closed-form expression for this integral?

$$\int_{0}^{\infty} \frac{1}{(x-\log x)^2}dx$$

Generalization of the Integral:

$$\int_{0}^{\infty} \frac{1}{(x-\log x)^{p}}dx$$

where, $\log x$ is a natural logarithm, $p\in\mathbb{Z^{+}}_{≥2}$
The indefinite integral can not be expressed by elementary mathematical functions according to Wolfram Alpha.
I can add a visual plot.

So, I dont know, is it possible to find a closed-form or not. But, I have a numerical solution:

$$\int_{0}^{\infty} \frac{1}{(x-\log x)^2}dx≈2.51792$$

 A: Surprise !
Considering
$$I=\int_{0}^{\infty} \frac{dx}{(x-\log x)^2}$$ and exploring simple linear combinations of a few basic constants, I found (be sure it took time !)
$$\color{blue}{I\sim\frac{189}{4}(C+2\pi)+61 \pi  \log (3)-\frac{1}{4} \left(57+101 \pi ^2+523 \pi  \log (2)\right)}$$ which differs in absolute value by $10^{-18}$.
Update
Funny is 
$$J=\int_{1}^{\infty} \frac{dx}{(x-\log x)^2}\sim \frac{717 \pi ^2-489 \pi-296 }{405 \pi ^2-420 \pi+112}$$which differs in absolute value by $3 \times 10^{-19}$.
$$K=\int_{0}^{1} \frac{dx}{(x-\log x)^2}\sim \frac{157 e^2+693 e-1394}{489 e^2-62e-859}$$which differs in absolute value by $ 10^{-20}$.
So, another formula
$$\color{blue}{I=J+K \sim \frac{157 e^2+693 e-1394}{489 e^2-62e-859}+\frac{717 \pi ^2-489 \pi-296 }{405 \pi ^2-420 \pi+112}}$$ which differs in absolute value by $3 \times 10^{-19}$.
A: To Yuriy S: the method is good, you just forgot to consider all branches of Lambert W, the solution to $ e^x -x =0 $ is $ x_n= - W_{n}(-1) $ for $ n \in \mathbb{Z}$ so the integral is
$$
2\pi i \sum_{n=0}^{\infty} \text{Res}\left( \frac{e^x}{\left( e^x -x\right )^2}, -W_n(-1)\right) = 2\pi i \sum_{n=0}^{\infty}  -\frac{W_n(-1)}{\left( 1+ W_n(-1)\right )^3}
$$
Mathematica code:
Abs[Sum[NResidue[E^x/(E^x - x)^2, {x, -ProductLog[n, -1]},WorkingPrecision -> 50], {n, 0, 10000}]*2*Pi]

Real part of the sum converges VERY slowly to 0.
Sorry can't comment
A: A more general relationship :
$$\int_1^\infty \frac{dx}{(x-\ln(x))^p}=\sum_{k=0}^\infty \frac{(p+k-1)!}{(p-1)!\:(k+p-1)^{k+1}}\qquad p> 1 \:,\: p \text{ integer.}$$
This could be extended to real $p>1$ thanks to the function $\Gamma$.
Or, on an equivalent form :
$$\int_1^\infty \frac{dx}{(x-\ln(x))^p}=\frac{1}{(p-1)!}\sum_{n=p-1}^\infty \frac{n!}{n^{n-p+2}}\qquad p> 1 \:,\: p \text{ integer.}$$
From this it is easy to find the already known cases $p=2$ and $p=3$.
Then other cases, for example :
$$\int_1^\infty \frac{dx}{(x-\ln(x))^4}=\frac{1}{6}\sum_{n=3}^\infty \frac{n!}{n^{n-2}}$$
I let you the pleasure to prove the above formulas. 
This is not too difficult in expending  $\frac{1}{(1-t)^p}$ to series of powers of $t$ with $t=\frac{\ln(x)}{x}$ and knowing that $\int_1^\infty\frac{(\ln(x))^k}{x^{p+k}}dx=\frac{k!}{(k+p-1)^{k+1}}$.
