Finding a specific 2-form for ${ \mathbb{R}^3 }$

Question

Let $x_1$, $x_2$, $x_3$ coordinates in ${\mathbb{R}^3 }$. I need to find a 2-form $\omega$, which satisfies
${ d \omega =d x_1 \land d x_2 \land d x_3}$
How should i do this? Also I would like to know, whether there is a 1-form ${\alpha}$ with ${ \omega=d \alpha }$.

Answer 1

How about $$\omega=\frac{1}{3}\left(x_1\ dx_2\wedge dx_3-x_2\ dx_1\wedge dx_3+x_3\ dx_1\wedge dx_2\right)?$$

To answer your second question, no. Otherwise, we'd have $$dx_1\wedge dx_2\wedge dx_3=d\omega=d(d\alpha)=0.$$