How to arrive at $\int x^2 \phi(x) \, \mathrm{dx} = \Phi(x) - x\phi(x) + C$ 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.
