Here is the question: Let $\omega = A dy\wedge dz + B dz \wedge dx + C dx \wedge dy$ in $\mathbf{R}^3$, and $d\omega = 0$. Denote \begin{eqnarray} \alpha = \int_0^1 tA(tx,ty,tz)dt\cdot(ydz-zdy)\\ +\int_0^1 tB(tx,ty,tz)dt\cdot(zdx-xdz)\\ +\int_0^1 tC(tx,ty,tz)dt\cdot(xdy-ydx). \end{eqnarray} Then show that $d\alpha = \omega$.
Just follow the definitions. Write $\alpha$ as $adx+bdy+cdz$, then show that $b_x-a_y=C$ etc. On the way, you will need to use that $d\omega=0$, ie $A_x+B_y+C_z=0$.