Is the function $e^{x^2/2} \Phi(x)$ monotone increasing?

We can write $h(x)=\frac 1{\sqrt{2\pi}}\int_{-\infty}^x \exp\left(\frac{x^2-y^2}2\right)dy$. Now put $t=x-y$. We get \begin{align} h(x)&=\frac 1{\sqrt{2\pi}}\int_0^{+\infty}\exp\left(\frac{x^2-(x-t)^2}2\right)dt\\\ &=\frac 1{\sqrt{2\pi}}\int_0^{+\infty}\exp\left(xt-\frac{t^2}2\right)dt. \end{align} We can differentiate under the integral thanks to the dominated convergence theorem. We get $$h'(x)=\frac 1{\sqrt{2\pi}}\int_0^{+\infty}t\exp\left(xt-\frac{t^2}2\right)dt\geq 0.$$ Added later: we don't need to diffentiate. If $x_1\leq x_2$ then for $t\geq 0$ we have $e^{tx_1}\leq e^{tx_2}$ therefore $ h(x_1)\leq h(x_2) $.

This is just an alternative argument to Davide's nice one.

First, note that $h' = e^{x^2/2}(x \Phi + \Phi')$.

Since $\Phi'' = -x \Phi'$, monotonicity of the integral yields $$ x \Phi(x) \geq \int_{-\infty}^x u \Phi'(u) \mathrm{d}u = -\Phi'(x). $$ So, $h' \geq 0$, and we are done.

I haven't actually done the computation, but it seems to me that integrating the $\Phi(x)$ term by parts ad nauseam, you get a nice power series for $h(x).$

EDIT @Davide's argument is obviously the complete answer to the question as asked, but just as a coda, the series for $h(x)$ is quite cute:

In the odd part, the coefficients of $x^{2k+1}$ is $1/p(k)$ where $p(k)$ is the product of the first $k$ odd integers, while in the even part, the coefficient of $x^{2k}$ is $\sqrt{\pi/2}/q(k),$ where $q(k)$ is the product of the first $k$ even integers.