Let me as precise as possible. $u$ is an $r-1$-form such that $du=\varphi$. In particular, $du$ is of pure bidegree $(p,q)$. We have decomposed $u=\sum_{i=0}^{r-1}u^{i,r-1-i}$, where $u^{i,r-1-i}$ is the $(i,r-1-i)$ component of $u$. Then, we calculate $\varphi=du=\partial u+\overline{\partial}u=\sum_{i=0}^{r-1}\partial u^{i,r-1-i}+\sum_{i=0}^{r-1}\overline{\partial} u^{i,r-1-i}$. Now, $\partial u^{i,r-1-i}$ is an $(i+1,r-1-i)$-form and $\overline{\partial}u^{i,r-1-i}$ is an $(i,r-i)$-form. Thus, the $(r,s)$-part of $du$ is given by $\partial u^{r-1,s}+\overline{\partial}u^{r,s-1}$. Since …

alright, second attempt: prove that a row of the cayley table of a group $G$ has no identical element.
Let the set $U = \{x_1, ..., x_n\}$ be a permutation of the group $G$ with $n$ non-identical elements, let $g \in G$ be an element of the group, and let the set $V = \{y_1, ..., y_k\}$ be s.t. it's ith element satisfies $y_i = gx_i$ for $U$'s ith element. By definition of a group, $g$ is invertible, and therefore there is a one to one correspondence between $V$ and $U$. Therefore, all elements of $U$ are non-identical.

Third attempt: prove that no element of the row of a group $G$'s caley table is identical.
Let $g$ be an element of the group $G = \{x_i\}$ with $n$ elements, let the function $g: G \to G$ be s.t. $g(x) = gx$, and let the elements of the row of a cayley table be contained by $\{g(x_i)\}$. By definition of a group, there is an inverse element $g^{-1}$ that satisfies $g^{-1}gx = x$, and therefore, $g^{-1}g(x) = x$, i.e., $g(x)$ is invertible. Therefore $g(x)$ is bijective, and each element of $\{g(x_i)\}$ has a unique correspondent in $\{x_i\}$ and vice-versa, so no element in $\{g(x_i)\}$ is…

Oh, Oh, these fuzzy images are really causing headaches!
I'm trying to work through the suggestions I have recieved (and perservering with Ted's strategy before asking for more help.)
So far I have gone from this:
$$
-i\hbar \int\limits_{-\infty}^{+\infty}{\frac{\partial}{\partial t}\Psi^*\left[\frac{\partial}{\partial x}\right]\Psi}dx.
$$
to this
$$
i\hbar
\left(
\int\limits_{-\infty}^{+\infty}{
\frac{-i\hbar}{2m} \dfrac{\partial^2 \Psi^*}{\partial x^2} \frac{\partial \Psi}{\partial x} + \frac{i}{\hbar}V \Psi^* \frac{\partial \Psi}{\partial x}