$d(I\omega) + I(d\omega) = \omega$ for differential forms 
Let $U$ be a convex connected and open set in $\mathbb{R}^n$, such that $0\in U$. For every $k$-differential form $\omega$, $$\omega =\sum_{i_1\lt\dots\lt i_k} c_{i_1,\dots,i_k}(x)dx{_{i_1}}\wedge\dots\wedge dx{_{i_k}},$$ we define a new $k-1$ differential form, $I\omega$ as follows - 
$$
\begin{split}
(I\omega)
 = \sum_{i_1\lt\dots\lt i_k}
   \sum_{j=1}^kx_{i_j}
   &\left( \int_0^1s^{k-1}c_{i_1,\dots,i_k}(sx)ds\right)\\
&
dx_{i_1}\wedge\dots dx_{i_{j-1}}\wedge dx_{i_{j+1}}\wedge\dots\wedge dx_{i_k}
\end{split}
$$
Prove  that $d(I\omega) + I(d\omega) = \omega$.

So I know that - 
$d\omega = \sum_{i_1\lt\dots\lt i_k} \sum_{j=1}^n\frac{\partial c_{i_1,\dots,i_k}(x)}{\partial x_j}dx_j\wedge dx_{i_1}\wedge\dots\wedge dx_{i_k}$
and therefore - 
$I(d\omega) =  \sum_{i_1\lt\dots\lt i_k} \sum_{j=1}^k x_{i_j}(\int_0^1\frac{\partial c_{i_1,\dots,i_k}(sx)}{\partial x_j}ds)dx_{i_1}\wedge\dots dx_{i_{j-1}}\wedge dx_{i_{j+1}}\wedge\dots\wedge dx_{i_k}$
All I need now is to calculate $d(I\omega)$, and put it into $d(I\omega) + I(d\omega)$ and show that it is equal to $\omega$.
However, I'm not sure - How can I calculate $d(I\omega)$?
What is $d(I\omega)=\sum_{i_1\lt\dots\lt i_k}x_{i_j}\sum_{j=1}^n\frac{\partial(\int_0^1s^{k-1}c_{i_1,\dots,i_k}(sx)ds)}{\partial d_{x_j}}dx_{i_1}\wedge\dots dx_{i_{j-1}}\wedge dx_{i_{j+1}}\wedge\dots\wedge dx_{i_k}$?
Or perhaps, is there any other way(some tricks) so I won't need to calculate it explicitly?
 A: Why I think you formula contains an error $\newcommand{\d}{\mathop{}\!d}$
Let $n = 2$ and  $\omega = \d x \wedge \d y$. Then according to your formula
$$I \omega = x \left(\int_0^1 s \d s \right) \d y + y \left( \int_0^1 s \d s\right) \d x = \frac12 (x \d y + y \d x) $$
so $\d (I \omega) = \frac12(\d x \wedge \d y + \d y \wedge \d x) = 0$ and since $\d\omega = 0$ the equation you want to prove does not hold. However if you add a factor of $(-1)^{j-1}$ to the inner sum in the definition of $I \omega$ then $I \omega$ would be $\frac12(x \d y - y \d x)$ in this example and everything works out. The formula for $I$ then is
$$
(I\omega)
 = \sum_{i_1\lt\dots\lt i_k}
   \sum_{j=1}^kx_{i_j} 
   \left( \int_0^1s^{k-1}c_{i_1,\dots,i_k}(sx) \d s\right)
(-1)^{j-1} \d x_{i_1}\wedge\dots \wedge \d x_{i_{j-1}}\wedge \d x_{i_{j+1}}\wedge\dots\wedge \d x_{i_k}
$$
Proving $\d(I \omega) + I(\d \omega) = \omega$
Since $\d$ and $I$ are bot additive, it is sufficient to verify this for $\omega = f(x) \d x_{i_1} \wedge \dots \wedge \d x_{i_k}$. By further reordering of the coordinates we can assume $\omega = f \d x_1 \wedge \dots \wedge d x_k$.
Let's denote $\d x_1 \wedge \dots \wedge d x_k$ by $\d x_I$ and $\d x_1 \wedge \dots \wedge \d x_{j-1} \wedge \d x_{j+1} \wedge \dots \wedge d x_k$ by $\d x_{I - j}$.
We have
$$ I \omega = \sum_{j=1}^k x_j \left( \int_0^1 s^{k-1} f(s x) \d s \right) (-1)^{j-1} \d x_{I - j}
$$
and 
$$
\begin{align}
\d (I \omega) &= \sum_{i = 1}^n \sum_{j=1}^k \frac{\partial}{\partial x_i} \left[x_j  \int_0^1 s^{k-1} f(s x) \d s  \right] \d x_i \wedge (-1)^{j-1} d x_{I - j} \\
&= \sum_{i=1}^k \frac{\partial}{\partial x_i} \left[x_i  \int_0^1 s^{k-1} f(s x) \d s  \right] \d x_I \\
&\phantom{=}+ \sum_{i = k + 1}^n \sum_{j=1}^k  x_j  \left( \int_0^1 s^k \frac{\partial f}{\partial x_i}(s x) \d s  \right) (-1)^{j-1} (-1)^{k-1} d x_{I - j} \wedge d x_i \tag{1}
\end{align}
$$
We can simplify the first sum:
$$
\begin{align}
\sum_{i=1}^k \frac{\partial}{\partial x_i} \left[x_i  \int_0^1 s^{k-1} f(s x) \d s  \right] &= 
        \int_0^1 k s^{k-1} f(s x) \d s + \int_0^1 s^k \sum_{i=1}^k x_i \frac{\partial f}{\partial x_i} (s x)
\end{align}
$$
Now
$$
\d \omega = \sum_{i = k + 1}^n \frac{\partial f}{\partial x_i} \d x_i \wedge \d x_I
          = (-1)^k \sum_{i = k + 1}^n \frac{\partial f}{\partial x_i} \d x_I \wedge \d x_i
$$
and
$$
\begin{align*}
 I(\d \omega) = (-1)^k \sum_{i = k+1}^n &\left[ \sum_{j=1}^k x_j \left( \int_0^1 s^k \frac{\partial f}{\partial x_i}(s x) \right) (-1)^{j-1} \d x_{I - j} \wedge \d x_i \right.\\
    &\phantom{[}\left. + x_i \left( \int_0^1 s^k \frac{\partial f}{\partial x_i}(s x) \right) (-1)^k \d x_I\right] \tag{2}
\end{align*}
$$
The first sum in $(2)$ is exactly the negative of the second sum in $(1)$ - $(-1)^k$ vs $(-1)^{k-1}$ - so they cancel when adding $I(\d \omega)$ and $\d (I \omega)$ What remains is
$$
\left( \int_0^1 k s^{k-1} f(s x) \d s + \int_0^1 s^k \sum_{i=1}^n x_i \frac{\partial f}{\partial x_i} (s x) \right) \d x_I
$$
Since the sum on the right hand side is the total derivative of $f(s x)$ with respect to $s$, partial integration shows that this is equal to $f(x) \d x_I = \omega$.
