= Solution
A <coorientation> gives a globally nonvanishing $C^2$ <one-form> $\zeta$ with $E=\ker\zeta$. For tangent fields $A,B$ in the <integrable distribution>,
$$
d\zeta(A,B)=-\zeta([A,B])=0,
$$
since integrability makes their bracket tangent to $E$. Thus $\zeta\wedge d\zeta=0$. Choose a transverse field $Y$ with $\zeta(Y)=1$. The $C^1$ <one-form> $\eta=-\iota_Yd\zeta$ then satisfies $d\zeta=\eta\wedge\zeta$, as is checked on $E$ and $Y$.
The <Godbillon-Vey invariant> in the unnormalized convention is
$$
\boxed{\mathrm{gv}(E)=\int_N\eta\wedge d\eta}.
$$
The three-form is continuous at the specified regularity, so its integral is defined. We prove independence of both choices by producing exact differences with $C^1$ primitives, allowing ordinary <Stokes theorem>.
First, for a fixed $\zeta$, another admissible auxiliary form has the form $\eta' =\eta+a\zeta$. From $d^2\zeta=0$ and $d\zeta=\eta\wedge\zeta$, one gets $d\eta\wedge\zeta=0$. Expanding the two three-forms, using this relation and $\zeta\wedge d\zeta=0$, gives
$$
\begin{aligned}
\eta'\wedge d\eta'-\eta\wedge d\eta
&=\eta\wedge da\wedge\zeta\\
&=-da\wedge d\zeta=-d(a\,d\zeta).
\end{aligned}
$$
Its integral on the closed manifold vanishes.
Next any other defining form is $\zeta'=h\zeta$ with $h$ nowhere zero. Let $g=\log|h|$. Then
$$
d\zeta'=(\eta+dg)\wedge\zeta',
$$
so one may use $\eta_0'=\eta+dg$. Its three-form differs by
$$
\eta_0'\wedge d\eta_0'-\eta\wedge d\eta
=dg\wedge d\eta=d(-dg\wedge\eta).
$$
Again the difference integrates to zero; any subsequent auxiliary choice is handled by the first calculation. This includes reversing the chosen <coorientation>, since $h$ can be negative. Thus the invariant, and the corresponding <de Rham cohomology> class, depends only on the oriented manifold and the <foliation> defined by $E$. Reversing the orientation of $N$, unlike reversing the <coorientation> of $E$, changes the sign of its numerical value.
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