Does this integral have a close form solution? Here is the question:

$$ \int_{0}^{1}\, \frac{1}{(x\mathrm{log}\, x+1)^{2}}dx $$

I have tried the residue to solve this by choosing a rectangle contour, but by doing this I obtained a series containing Lambert-W funtion I cannot work out. I guess there is no close-form solution to this integral.
Could you please tell me if it has a close-form solution? Thanks in advance!
 A: I should be really surprized by a closed form for this integral.
You can approximate it using $y=x \log(x)$ which makes 
$$ \frac{1}{(x\mathrm{log}\, x+1)^{2}}=\frac 1{(y+1)^2}$$ and expand it as a Taylor series around $y=0$. This will give 
$$\frac 1{(y+1)^2}=\sum_{n=1}^p (1)^{n-1} n y^{n-1}$$ and we are then let with $$I_n=\int_0^1 x^n \log^n(x)=e^{i \pi  n} (n+1)^{-n-1} \Gamma (n+1)$$ which is valid for any value of $n$.
This makes, for integer $n$,
$$\int_{0}^{1}\, \frac{dx}{(x\mathrm{log}\, x+1)^{2}}=\sum_{n=1}^p   \frac{\Gamma (n)}{n^{n-1}}$$ which seems to converge quite quickly
$$\left(
\begin{array}{cc}
 p & \sum_{n=1}^p \frac{\Gamma (n)}{n^{n-1}} \\
 5 & 1.8543722 \\
 6 & 1.8698043 \\
 7 & 1.8759242 \\
 8 & 1.8783275 \\
 9 & 1.8792641 \\
 10 & 1.8796270 \\
 11 & 1.8797669 \\
 12 & 1.8798206 \\
 13 & 1.8798412 \\
 14 & 1.8798491 \\
 15 & 1.8798520
\end{array}
\right)$$
This should be easy to program as $$\sum_{i=1}^p u_i \qquad \text{with}\qquad a_1=1 \qquad \text{and}\qquad u_{n+1}=\left(\frac{n}{n+1}\right)^n\, u_n$$
