I am self-studying Normal approximations with Malliavin calculus by Ivan Nourdin and Giovanni Peccati (btw I think the book is marvelous).
After having defined $\gamma$ as the standard Gaussian probability measure and having taken $f,g \in C^{\infty}$ Nourdin and Peccati state in Lemma 1.1.6 that
$$\int_R f^{(p)}(x) g(x) d \gamma(x) = \int_R f^{(p-1)}(x) (xg(x) - g'(x)) d \gamma(x)$$
this, they say, should follow from lemma 1.1.1 that says that letting $f: R \rightarrow R$ be an absolutely continuous function such that $f' \in L^1( \gamma)$. Then $x \rightarrow x f(x) \in L^1(\gamma)$ and $$ \int_R x f(x) d \gamma(x) = \int_R f'(x) d \gamma(x) $$.
I can't see why this would be, my calculations get me close but something is missing:
$$\int_R f^{(p)}(x) g(x) dx = g(x)f^{(p-1)}(x) - \int_R f^{(p-1)}(x) g'(x) d x$$
and if $\int_R f^{(p-1)}(x) g'(x) d x = f^{(p-1)}(x) g'(x) /x$ I would be finished but this is not the case in general.