Q:

sufficient condition for being an integral factor

Let $ f: \mathbb {R}^m \rightarrow \mathbb {R}-\{0\} $ function $C^{\infty}$ class and $w$ a one-form $C^{\infty}$ class in $\mathbb {R}^m $. 
If $\alpha=w-\dfrac{1}{f}dx_{m+1} $ satisfies $\alpha \wedge d\alpha= w \wedge dw$ then $d(fw)=0$
Note: $\mathbb {R}^m \subseteq \mathbb {R}^{m+1}$ with $x_{m+1}=0$.
Thanks for any suggestions.

A:

Well, by a straightforward computation,
$$
 \alpha \wedge d\alpha = (\omega - f^{-1}dx^{m+1}) \wedge d (\omega - f^{-1}dx^{m+1})\\ = (\omega - f^{-1}dx^{m+1}) \wedge (d\omega + f^{-2} df \wedge dx^{m+1})\\ = \omega \wedge d \omega + \omega \wedge f^{-2}df \wedge dx^{m+1} - f^{-1}dx^{m+1} \wedge d\omega\\ = \omega  \wedge d\omega - f^{-2} (df \wedge \omega \wedge dx^{m+1} + f d\omega \wedge dx^{m+1})\\
= \omega \wedge d\omega - f^{-2} d(f \omega ) \wedge dx^{m+1}.
$$
Now, given what you know, what can you conclude about $d(f\omega)$?

