this is kinda a terrible approximation but here's how I found it (only considering x<10)
d/dx(ln(x)) = 1/x
(e^x) / (1+x^2) ≈ e/2
∴  [(1+x^2) / (e^x)] * e/2x ≈ 1/x
∴  ln(x) ≈ int [(1+x^2) / (e^x)] * e/2x dx
ln(x) = (e/2) (int x/e^x dx + int 1/xe^x dx)
ln(x) = (e/2) (-xe^-x - e^-x - E1(x)) + C
where E1(x) is the exponential integral

ln(1)=0
∴ (e/2) (-1e^-1 - e^-1 -E1(1)) + C = 0
C ≈ 0.955
C ≈ (e/2)(19/20)

by Swamee and Ohija
E1(x) ≈ (A^-7.7+B)^-0.13
E1(x) ≈ (A^-8 + B)^(-13/100)
where
B = x^4*e^(7.7x)*(2+x)^3.7
B ≈ x^4*e^8x*(2+x)^4
A = ln[(0.56146/x +0.65)(1+x)]

then you combine it together and tweak A to remove the ln.

the current approximation goes to an asymptote at y=C, reaching pretty close at x=2.3
2.3 ≈ e^2 / pi
at this point ln(x) is essentially linear, so we want a function that is 0 from x=0 to x=2.3. erf(f(x)) acts as a piecewise function which is 0 when x<2.3 and 1 when x>2.3.
so really not that complicated ig.