$dx$ being a desginator (with respect to $x$) or being a term? I am confused as to what $dx$ truly is. I am doing some u-substitution problems and this is what I came across:
$$\int 2x(x-1)^{1/2}\,dx$$
$u=x-1$ and therefore $du=1$
when we substitute we get:
$$2 \int (u^{3/2}+u^{1/2}) \, du$$
(here the du simply replaces the dx because our variable changed)
In another example:
$$\int 4x^5(x^2+1)^{1/3} \, dx$$
$$u=x^3+1$$
$$du=3x^2$$
therefore it becomes:
$$\int 4(u-1)(du/3)(u)^{1/3}\,$$
-here my teacher didn't put $d$x at the end, she just left it off
So my question is this: why is it that sometimes $dx$ and $du$ are treated as values that can be multiplied to other terms in the integrand and sometimes they are simply treated as a command (do "blank" with respect to $x$, or $u$ or whatever is used)? 

Consider
$$\begin{array}{lll}
\displaystyle\int 4x^5(x^2+1)^\frac{1}{3}dx&=&\displaystyle\int 4x^5(x^2+1)^\frac{1}{3}dx\\
&=&\displaystyle2\int (x^2)^2(2x)(x^2+1)^\frac{1}{3}dx\\
&=&\displaystyle2\int ((x^2+1)-1)^2(2x)(x^2+1)^\frac{1}{3}dx\\
\end{array}$$
Next, let's make the substitution
$u=x^2+1$, implying that $du/dx=2x$.
We do this by multiplying by $1$.
$$\begin{array}{lll}
&=&\displaystyle2\int ((x^2+1)-1)^2(2x)(x^2+1)^\frac{1}{3}dx\cdot\frac{\frac{du}{dx}}{\frac{du}{dx}}\\
&=&\displaystyle2\int \frac{((x^2+1)-1)^2(2x)(x^2+1)^\frac{1}{3}\frac{dx\cdot du}{dx}}{\frac{du}{dx}}\\
&=&\displaystyle2\int \frac{((x^2+1)-1)^2(2x)(x^2+1)^\frac{1}{3}du}{2x}\\
\end{array}$$
Notice how I interpreted $du/dx$ as a ratio of differential in the numerator, but as a derivative in the numerator. This "inconsistency" isn't a bad thing, for I think that everyone can remember in middle/high school how useful it was to interpret a division as a multiplication (e.g. $5\div2=5\times\frac{1}{2}$) while manipulating fractions. Extending this technique of "multiple interpretations" to calculus, can only make your mathematical skills stronger.
Continuing
$$\begin{array}{lll}
&=&\displaystyle2\int \frac{((x^2+1)-1)^2(2x)(x^2+1)^\frac{1}{3}du}{2x}\\
&=&\displaystyle2\int ((x^2+1)-1)^2(x^2+1)^\frac{1}{3}du\\
&=&\displaystyle2\int (u-1)^2u^\frac{1}{3}du\\
&=&\dots\\
\end{array}$$
