Let $X \sim Normal(0,\sigma^2)$ and $g:\mathbb{R}\rightarrow\mathbb{R}$ a differentiable function such that there exists a positive integer $k$ such that
$$\lim_{|x|\to\infty} \Big|\frac{g(x)}{x^k}\Big|=0$$
Show that
$$\mathbb{E}[X \ g(X)]=\mathbb{E}[g'(X)]\mathbb{E}[X^2]$$
Let $f(x)=\dfrac{1}{\sigma\sqrt{2\pi}}e^{-\frac{x^2}{2\sigma^2}}$. Note that $\mathbb E[X^2]=\sigma^2$. Then integration by parts leads to $$ \mathbb{E}[g'(X)]\mathbb{E}[X^2] = \sigma^2 \int_{\mathbb R} g'(x) f(x)dx = \sigma^2 \int_{\mathbb R} f(x)dg(x) $$ $$ =\left[\sigma^2 f(x)g(x)\right]_{-\infty}^{+\infty} - \sigma^2 \int_{\mathbb R} g(x)\,df(x). $$ The first term is zero since $f(x)=o(|x|^{-k})$ as $|x|\to\infty$. And $df(x)=-xf(x)\,dx/\sigma^2$. Then the last term is exactly $\mathbb E[X\,g(X)]$.