Wolfram Alpha knows that, and it can be calculated that: $$ \int_0^1 \exp\bigg(\frac{1}{\log x}\bigg)~dx=2K_1(2).$$
Where $K_1$ is a modified Bessel function of the second kind.
I wanted to find out if there were closed forms when the upper bound was say $x=1/e$ or something less than $x=1.$
Wolfram Alpha was not able to find a closed form for this...
So, I sliced the integrand function by $y=x$ to halve the area, getting $K_1(2).$ Then I integrated $x,$ from $x=0$ to $x=1/e$ to get $\frac{1}{2e^2}.$ Next, I had the equation $\frac{1}{2e^2}+A=K_1(2).$ It follows that $A=K_1(2)-\frac{1}{2e^2}.$ By symmetry, $\int_0^{1/e} \exp\bigg(\frac{1}{\log x}\bigg)~dx=K_1(2)+\frac{1}{2e^2}.$
Are there closed forms for other upper bounds?
I don't think my method works for other upper bounds.
Only series.
With help of Wolfram Alpha:
$$\int_0^x \exp \left(\frac{1}{\log (t)}\right) \, dt=\\\int_0^x \left(\sum _{j=0}^{\infty } \frac{\left(\frac{1}{\log (t)}\right)^j}{\Gamma (j+1)}\right) \, dt=\\\sum _{j=0}^{\infty } \int_0^x \frac{\left(\frac{1}{\log (t)}\right)^j}{\Gamma (j+1)} \, dt=\\\sum _{j=0}^{\infty } \frac{(-1)^j \Gamma (1-j,-\log (x))}{\Gamma (j+1)}$$ if: $0<x<1$