Definite Integral in the study of Prophet Inequalities I wish to integrate the function $\frac{1}{a + x - x \cdot \ln x}$ for $x$ going from $0$ to $1$, where $0<a<1$ is a constant. It's easy to see that when $a=0$ this function has a simple indefinite integral $-\ln(\ln x - 1)$. However, for non-zero $a$, solvers like Mathematica are failing to find the indefinite integral.
More particularly, for my work I am interested in finding $a$ that is the solution to the following equation: 
$$ \int_{x=0}^{1} \frac{1}{a + x - x \cdot \ln x} dx = 1. $$
One can use numerical calculators to check that $a$ is close to $0.34148$, but I am hoping to obtain a "closed-form'' solution or find out if this cannot be simplified further. Thanks!
Remark: This integral appears in the study of Prophet Inequalities. See the following paper "Comparisons of Stop Rule and Supremum Expectations of I.I.D. Random Variables" of Hill and Kertz: http://www.jstor.org/stable/2243434

The bad news is that the integral actually diverges at $0,$ despite the nice indefinite integral. The good news is that if you evaluate it (say, numerically) at one point ($a=1$ is good), you have a rapidly converging power series expansion. Namely, by differentiating Brevan's expression under the integral sign with respect to $a.$ Actually, you dont even have to do that. Writing Brevan's integrand as 
$$\frac1 a \exp(-u+1) \frac{1}{1  + u\exp(-u+1)/a},$$ expand this in a geometric series, and note that each term is easily evaluated (and decays exponentially).
