indefinite integral of $e^{\frac{-x^2}{1+x}}$ I'm wondering if there's some simple form of the integral: $\int_0^X e^{\frac{-x^2}{1+x}} dx$
For context, I'm fitting to a distribution (of the velocities of local stars) as shown below and it's well approximated by $p(v) \propto e^\frac{-v^2}{1+|v|}$. AFAIK this is not a common probability distribution, but it seems a useful one, especially if the CDF can be written in simple functions.
histogram of observed stars in red, and the proposed pdf in green
 A: It doesn't look like there is an elementary integral, but we can find asymptotic forms.
For small $X$
Since $\frac{-x^2}{1+x}=1-x-\frac{1}{1+x}$, I will consider the integral
$$
I(b)=\int\limits_0^b dx \  e^{ -x-\frac{1}{1+x}}
$$
Which differs from the original by a multiplicative constant $e$, and investigate $b \to 0$. Integrating by parts
$$
I(b)=\int\limits_0^b dx \ (1+x)^2 e^{-x} \left[ \frac{d}{dx} e^{-\frac{1}{1+x}}\right] 
$$
$$
I(b)=(1+x)^2  e^{ -x-\frac{1}{1+x}} \Bigg\vert_0^b - \int\limits_0^b dx \ (1-x^2) e^{ -x-\frac{1}{1+x}}
$$
The boundary term may be readily evaluated. Within the integral on the right we recognize $I(b)$
$$
2 I(b)=(1+b)^2 e^{ -b-\frac{1}{1+b}}-e^{-1}+\int\limits_0^bx^2e^{ -x-\frac{1}{1+x}}
$$
For $b \to 0$, the integral on the right is $O(b^3)$ and we have
$$
I(b)>>\frac{1}{2} \int\limits_0^bx^2e^{ -x-\frac{1}{1+x}}  \ \ , \ \ b \to 0
$$
Therefore
$$
I(b) \sim \frac{1}{2} \left( (1+b)^2 e^{ -b-\frac{1}{1+b}}-e^{-1} \right) \ \ , \ \ b \to 0
$$
Simplifying, I find the leading order term
$$
I(b) \sim \frac{b}{e} \ \ , \ \ b \to 0
$$
In principle, repeated integration by parts could produce the asymptotic series. Here is a plot of the first term versus the numerical integration for small $b$:

For large $X$
I'll use the substitution in the comments: $t=x+1$ to write
$$
e^{-2} \int\limits_0^X \ dx \exp \left(-\frac{x^2}{1+x} \right) =\int\limits_1^a dt \ \exp(-t-t^{-1})
$$
And we're investigating $a \to \infty$. Let $f(t)=\exp(-t-t^{-1})$, rewrite the integral
$$
\int\limits_0^\infty dt \ f(t) = \int\limits_0^1 dt \ f(t) + \int\limits_1^a dt \ f(t) + \int\limits_a^\infty dt \ f(t)
$$
The integral on the left is exactly $2K_1(2)$, where $K$ is a modified Bessel function of the second kind. The first integral on the right is also constant, independent of $a$. We are left with the simpler problem of studying
$$
I(a)=\int\limits_a^\infty dt \ e^{-t-t^{-1}}
$$
For large $a$. Integrate by parts
$$
I(a)=\int\limits_a^\infty dt \ e^{-t^{-1}} \frac{d}{dt} \left[ -e^{-t} \right]
$$
$$
I(a)=-e^{-t-t^{-1}} \Big\vert_a^\infty + \int\limits_a^\infty dt \ \frac{1}{t^2} e^{-t-t^{-1}}
$$
The integral on the right differs from $I(a)$ by a factor of $1/t^2$, thus we have
$$
I(a)>> \int\limits_a^\infty dt \ \frac{1}{t^2} e^{-t-t^{-1}} \ \ , \ \ a \to \infty 
$$
Which leads to
$$
I(a) \sim e^{-a} \ \ , \ \ a \to \infty 
$$
Finally,
$$
\int\limits_1^a dt \ \exp(-t-t^{-1}) \sim 2K_1(2) - C - e^{-a} \ \ , \ \ a \to \infty
$$
Where $C=\int_0^1 dt \ f(t)$, and can be found numerically to be about $0.072$; probably there is a nice way to estimate it, but I don't see it right now. Here is a plot of the approximation versus the exact numerical integral:

EDIT: In terms of the incomplete Bessel functions, defined as
$$
K_\nu(x,y)=\int\limits_1^\infty dt \ t^{-\nu-1} \exp(-xt -y/t)
$$
We have for $I(a)$, by changing variables $u = t/a$
$$
I(a)=a \int\limits_1^\infty du \  e^{-au-a^{-1} u^{-1}} = a K_{-1}(a,a^{-1})
$$
And for $C$, by changing variables $u= 1/t$
$$
C= \int\limits_1^\infty du \ u^{-2} e^{-u- u^{-1}} =  K_{1}(1,1)
$$
Thus your original integral, with $a=X+1$ may be written
$$
\int\limits_0^X \ dx \exp \left(-\frac{x^2}{1+x} \right) = e^2 \left[ 2K_1(2) - K_{1}(1,1) - a K_{-1}(a,a^{-1}) \right] 
$$
