TITLE: What does it mean when dx is put on the start in an integral?
QUESTION [22 upvotes]: I have seen something like this before: $\int \frac{dx}{(e+1)^2}$. This is apparently another way to write $\int \frac{1}{(e+1)^2}dx$.
However, considering this statement: $\int\frac{du}{(u-1)u^2} = \int du(\frac{1}{u-1}-\frac{1}{u}-\frac{1}{u^2})$. On the left side, $du$ is moved, If I had to evaluate an integral that is written in this way, how would I expand it into the usual $\int f(x)dx$ form?
(From the comments) Is this truly a product and if not why is it commutative?

REPLY [3 votes]: I think as others are getting at, the confusion probably arises from the (great) question: what is $dx$ anyway? I will try and give an intuitive answer that should help a little with the confusion.
The rigorous answer is that $dx$ is a differential 1-form, which you will learn about if you do real analysis, but I think a perfectly good intuitive response is it is just a "tiny bit" of $x$. You can see this by noting that the integral is roughly defined as
$$
\lim_{\Delta x\rightarrow 0}\sum f(x*)\Delta x
$$
For some $x*$ inside each interval of shrinking size. $dx$ is then exactly this arbitrarily small $\Delta x$. As such, it is just an object that commutes with multiplication like $\Delta x$, and can be moved around as you wish. 
Thinking in this way, your question is the same as
$$
\sum f(x*)\Delta x=\sum \Delta xf(x*)
$$ 
where I hope you will believe me it is true. 
Additionally, I think this intuition for differentials will be helpful when you get to multi-variable calculus and define the volume element
$$
\mathrm dv=\mathrm dx \mathrm dy \mathrm dz
$$
which can be thought of as a tiny cube.