If $x\sim\mathcal{N}(\mu, \sigma^2)$, what is $E\left(\frac{\exp(x)}{a+\exp(x)}\right)$ for $a>0$? Let $x\sim\mathcal{N}(\mu, \sigma^2)$ be a normal random variable. I'm interested in a new random variable, defined as 
$$
z = \frac{\exp(x)}{a+\exp(x)},\quad x\in\mathbb{R}, a&gt;0.
$$
More specifically, I would like to find the closed form (if any) of its expected value, $\mathbb{E}[z]$, but I don't know how to proceed.

As a first step I could use an intermediate random variable $y=\exp(x)$, which follows the log-normal distribution (since $x$ is normal) and its mean and variance are given w.r.t. to $x$'s mean and variance. But again, I'm not sure if that helps.
If we apply a random variable transformation $Z = g(X)$ 

$f_Z(z)=f_X(g^{-1}(z))|\frac{d}{dz}g^{-1}(z)|$

where $g(X)=\frac{\exp[X]}{a+ \exp[X]}$, $f_X(x) = \frac{1}{\sqrt{2 \, \pi} \, \sigma}\exp[-\frac{(x-\mu)^2}{2\sigma^2}]$, and $g^{-1}(z)=\log\left(\frac{1}{\frac{1}{a\,z}-\frac{1}{a}}\right)$

$f_Z(z)= \frac{e^{-\frac{\left(\mu -\log \left(-\frac{a z}{z-1}\right)\right)^2}{2 \sigma
   ^2}}}{\sqrt{2 \pi } \sigma  \left| (z-1) z\right| }$

Finally

$E[Z]=\int z\,f_Z(z)dz = \frac{\int \frac{z e^{-\frac{\left(\mu -\log \left(-\frac{a z}{z-1}\right)\right)^2}{2
   \sigma ^2}}}{\left| (z-1) z\right| } \, dz}{\sqrt{2 \pi } \sigma }$
$F_Z(z)=P(Z\le z)=P(\frac{1}{1+ae^{-X}}\le z) = P(Ln(\frac{az}{1-z}) \le X) = F_X(Ln(\frac{az}{1-z}) )$

So $f_Z(z)=\frac{\partial}{\partial z}F_X(Ln(\frac{az}{1-z}) )  = \frac{1}{z(1-z)} f_X(Ln(\frac{az}{1-z}))$. Where $f_X$ is the usual normal density function. 
I don't think this will leave you much wiser so better to go after $E[Z]$ directly.