I found the following result in Wikipedia
$\int x^2 \phi(x) \, \mathrm{dx} = \Phi(x) - x\phi(x) + C$
where $$\phi(x) = \frac{1}{(2\pi)^{1/2}} \mathrm{e}^{-\frac{1}{2}x^2}~\mathrm{and}~\Phi(x) = \int_{-\infty}^{x} \phi(\mathrm{t}) \mathrm{dt}\,.$$
I tried some partial integration so far but somehow I don't get it. Can someone tell me how to arrive at the equality?
If we let $f(x) := \Phi(x) - x\phi(x)$, we have - using that $\phi'(x) = -x\phi(x)$ - \begin{align*} f'(x) &= \Phi'(x) - \phi(x) - x\phi'(x)\\ &= -x\phi'(x)\\ &= x^2 \phi(x) \end{align*} That is $$ \int x^2\phi(x) \, dx = f(x) = \Phi(x) - x\phi(x) $$ by the fundamental theorem of calculus.