Say I have an expression like this:
$$\partial C/\partial a = \frac{(a - y)}{a(1-a)x}$$
where $a$ and $y$ are independent, but $a$ is a function of $x$ and possibly some other variables (i.e. $a = f(x)$). Does it make sense to integrate this expression wrt to $a$, even though $a$ depends on $x$? If it is possible, how would it be done for this example?
Yes, it makes sense — but our modern calculus notation doesn't always make this clear. Let me try a variant notation to help.
Let's call $C$ a function of three variables $C(a,x,y)$. These variables are all independent. It makes sense to take partial derivatives with respect to any of these three arguments. We can call the partial derivatives $\partial_1 C$, $\partial_2 C$, $\partial_3 C$.
Now suppose we want to make some of these variables depend on each other. For example, we might want to make $a$ a function of $x$ --- so that $a = f(x)$. In this case, the independent variables will be $x$ and $y$, instead of $x$, $y$, and $a$.
We will need a function which explains how to convert our new independent variables into our old independent variables. In this case, the conversion function is:
$$T(x,y) = \langle f(x),\, x,\, y\rangle$$
And the updated version of $C$ is a new function:
$$C_\text{new} = C \circ T.$$
The definition here is just shorthand for saying that
$$C_{\text{new}}(x,y) = C(f(x), x, y).$$
Now, $C_\text{new}$ is, by design, a function with only two arguments. It makes sense to take partial derivatives with respect to either argument. We can call the partial derivatives $\partial_1 C_{\text{new}}$ and $\partial_2 C_{\text{new}}$.
In this new notation, the derivative you originally stated $\left(\frac{\partial C}{\partial a}\right)$ is now more clearly expressed as $\partial_1 C_{\text{new}}$. And you can easily evaluate this derivative using the chain rule, or integrate it.