I am trying to calculate $$\int e^{x-1} \frac{1}{x} dx $$ but got stuck. I first used substitution to rewrite the integral as $$\int e^{x} \frac{1}{x+1} dx .$$ Then I made the substitution $\sqrt{x} = \log(t)$, which gives $\frac{1}{x+1} = \frac{1}{\log(t)^2 + 1}$ and $dx = 2\log(t) \frac{1}{t} dt$, so the integral is equivalent to $$ \int \frac{2\log(t)dt}{(\log(t))^2 + 1}$$ Here I got stuck, because making the logical substitution of $s = \log(t)^2 + 1$ just results in the integral $\int e^{\sqrt{s-1}} \cdot \frac{1}{s} ds$, the integral I started with.
I am really stuck at this point and would appreciate any hints or advice to finding a solution.
This is the well-known exponential integral function (to a factor $e$).
The simple fact that a special function has been defined for this purpose is enough to say that there is no other closed form.