How to compute the following formula?
$$ \int_{-\infty}^{+\infty} \Phi(x) N(x\mid\mu,\sigma^2) \, dx $$
$$ \int_{-\infty}^{+\infty} \Phi(x) N(x\mid\mu,\sigma^2) x\,dx $$
where $\Phi(x)=\int_{-\infty}^{x}\frac{1}{\sqrt{2\pi}} \exp{(-t^2/2)} \, dt$, namely, the cumulative distribution function of normal distribution $N(0,1)$.
$N(x\mid\mu,\sigma^2) $ means the probability density function of Gaussian distribution with mean $\mu$ and variance $\sigma^2$
The first integral is solved in Distribution of the normal cdf.
So we know that $$ \int_{-\infty}^{\infty} \Phi \left (x \right ) N \left ( x|\mu,\sigma^2 \right ) dx = \Phi \left (\frac{\mu}{\sqrt{1+\sigma^2}} \right ) $$
Now, if we differentiate both sides w.r.t. $\mu$, $$\int_{-\infty}^{\infty} \frac{x-\mu}{\sigma^2} \Phi \left (x \right ) N \left ( x|\mu,\sigma^2 \right ) dx = \frac{1}{\sqrt{1+\sigma^2}} N \left (\frac{\mu}{\sqrt{1+\sigma^2}} \right ) $$
Notice the first term on the left hand side is what we want. Rearranging, $$\int_{-\infty}^{\infty} \Phi \left (x \right ) N \left ( x|\mu,\sigma^2 \right ) x dx = \mu \Phi \left ( \frac{\mu}{\sqrt{1+\sigma^2}} \right ) + \frac{\sigma^2}{\sqrt{1+\sigma^2}} N \left (\frac{\mu}{\sqrt{1+\sigma^2}} \right ) $$