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?
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First, note that $$ \frac{1}{u-1}-\frac{1}{u}-\frac{1}{u^{2}}=\frac{1}{u-1}-\frac{u+1}{u^{2}}=\frac{1}{\left(u-1\right)u^{2}} $$ from which the equality follows: $$ \int du \left(\frac{1}{\left(u-1\right)u^2}\right)=\int du\left(\frac{1}{u-1}-\frac{1}{u}-\frac{1}{u^{2}}\right). $$ Note that $\int fdu$ and $\int du f$ mean the same thing.
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While, as other answers have noted, some authors write the differential (the $du$) next to the integral sign, I would recommend avoiding it. It's uncommon, meaning many people who see it will have the same confusion you did. But also, having the differential at the end of an integral is a notational convenience, as it shows the end of the integrand expression. This is useful, as the integral notation doesn't inherently have a closing delimiter, other than the differential.
For instance, one could easily mistake
$$\int du\left(\frac{1}{u-1}-\frac{1}{u}-\frac{1}{u^2}\right)$$
as
$$\left(\frac{1}{u-1}-\frac{1}{u}-\frac{1}{u^2}\right)\int 1\, du,$$
which is a completely different expression. By writing
$$\int\left(\frac{1}{u-1}-\frac{1}{u}-\frac{1}{u^2}\right)\,du$$
it is clear what the integrand is. Indeed, even if the parentheses are omitted it is clear:
$$\int\frac{1}{u-1}-\frac{1}{u}-\frac{1}{u^2}\,du$$
The exception is that for integrands that are a fraction with numerator 1, they are often written as $$\int\frac{dx}{f(x)}$$ instead of $$\int\frac{1}{f(x)}\,dx,$$ since the former is more compact, and the fraction bar delimits the entirety of the integral expression.
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Others have explained the syntax and semantics of this notation, but no one seems to have addressed why it is a good thing to do...
The notation "$\dfrac{\mathrm{d}}{\mathrm{d}u}$" is used for the differentiation operator (a gadget which takes functions as input and produces functions as output). This is the "differentiate with respect to $u$ operator". Typical uses: \begin{align} \dfrac{\mathrm{d}}{\mathrm{d}x}&(x^2) = 2x \\ \dfrac{\mathrm{d}}{\mathrm{d}x}&\,x^2 = 2x \\ \dfrac{\mathrm{d}}{\mathrm{d}x}&\,f(x) = f'(x) \\ \dfrac{\mathrm{d}}{\mathrm{d}x}&\,f(u) = 0 \\ \dfrac{\mathrm{d}x^2}{\mathrm{d}x}& = 2x \text{.} \end{align} Except for the last example usage, these read left-to-right as "the derivative with respect to $x$ of ...". (This is unfortunate because the last form is the one that most visibly echoes a difference quotient.) So you know at the outset which variable is being varied. This has the downside of not unambiguously indicating where the argument to the operator ends (but only in the second, third, and fourth, and sadly most common, forms above).
The notation "$\int \mathrm{d}u$" is used for the antidifferentiation operator. (And the obvious extension is used for the integration operation.) This is the "antidifferentiate with respect to $u$ operator". Typical uses: \begin{align} \int \mathrm{d}x (2x) = x^2 + C \\ \int \mathrm{d}x 2x = x^2 + C \\ \int 2x \,\mathrm{d}x = x^2 + C \text{.} \end{align}
Similar comments apply as above: Except for the last example usage, these read as "the antiderivative with respect to $x$ of ...". That is, you know which variable is the variable of integration before you see any part of the argument of the operator. The second form has the same defect as the most common differentation form: it is unclear where the argument of the operator ends. The last form inserts the argument into the middle of the operator, so the entire argument is recited before one knows which operator is acting on it, but has the advantage of echoing the form of the Riemann sum (more relevant to integrals than to antiderivatives) and indicating where the argument ends.
You will also see this notation in iterated integrals. When this notation is used, it is to remind the reader that the integrals are iterated: $\int_a^b \mathrm{d}x \, \int_c^d \mathrm{d}y \ x + y^2$ says to pass $x+y^2$ through the "integrate with respect to $y$ on $[c,d]$" operation before passing the result through the "integrate with respect to $x$ on $[a,b]$" operation. (Note that the closedness and openness of the endpoints of those intervals is irrelevant.) This may seem pedantic until one learns that the value of an iterated integral can depend on the order in which the integrals are taken and the operator "$\int \int \mathrm{d}x \, \mathrm{d}y$" is only defined if it does not matter in which order the integration occurs.
Further, for iterated antiderivatives, the order seriously matters, as becomes evident in solving PDEs: \begin{align} \int \mathrm{d}x &\int \mathrm{d}y (x+y^2) \\ &= \int \mathrm{d}x (x y + \frac{y^3}{3} + C_1(x)) \\ &= \frac{x^2 y}{2} + \frac{x y^3}{3} + C_2(y) + \int \mathrm{d}x(C_1(x)) \text{,} \end{align} where $C_1(x)$ and $C_2(y)$ are arbitrary functions of $x$ and $y$, respectively. ($C_1$ comes in because $$\dfrac{\mathrm{d}}{\mathrm{d}y} \left( x y + \frac{y^3}{3} + C_1(x) \right) = x+y^2 + 0 = x+y^2 \text{.} $$ $C_2$ is an arbitrary function of $y$ for the analogous reason.) However, if we reverse the order of antidifferentiateion, the remaining integrated function depends only on $y$, not on $x$. Since not every arbitrary function is integrable, the two resulting solution sets need not be the same. Order seriously matters.
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In plain language, you can interpret the $\int$ and $\mathrm{d}$ as a verb or a command: "integrate" or "differentiate" (the following) and the $\mathrm{d}x$ as a complement to the verb: "with respect to variable $x$".
With $\dfrac{\mathrm{d}}{\mathrm{d}x}$, the verb and its complement are generally close by, and the fractional separation makes it quite obvious. Putting the $\mathrm{d}x$ at the end of the formula is like a sentence with the subject and the verb far away, giving a sense of surprise, yet not handy for the beginner. This sometimes happens with the German language for instance. Putting the $\mathrm{d}x$ close to the $\int$ sign may help the reader for long expressions.
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The hyperreal view helps clarify the picture with regard to the question whether $f(x)dx$ in the expression $\int_a^b f(x)dx$ is a product or not, as well as the issue of commutativity.
From the hyperreal viewpoint, the integral is defined as the standard part of an infinite Riemann sum $\sum_i f(x_i)\Delta x$. In the Riemann sum, the term $f(x_i)\Delta x$ is literally a product, and therefore commutative. Since the infinite Riemann sum uniquely determines the integral (via applying standard part), it follows that in the integral also we are free to commute $f(x)$ and $dx$.
Thus in the integral the term $f(x)dx$ is not literally a product, but the commutation property follows from that for Riemann sums via transfer.
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An operator is an object that acts on another object, typically to the right, though proper definition of the space you're working in will clarify that. We require that operators and the objects they act on fulfill certain requirements. Integrals and derivatives acting on functions fulfill the requirements for a vector space of functions. When an integral is written in operator notation, $\int dx \, f(x)$, assume that it could be equivalently written as $\mathbf{A \, f}$. Operator notation is useful because it transforms calculus into linear algebra for a little while, and occasionally provides an end-run around having to do difficult calculations.
For example, suppose we define two operators, $\mathbf{A} = \int dx$ and $\mathbf{B} = \int dy$ and some function $\mathbf{f} = f(x,y)$. While $\mathbf{A \, B \, f} \neq \mathbf{B \, A \, f}$ in general, if we define an object called the commutator as $[\mathbf{A,B}] ≡ \mathbf{AB - BA} $, we may write $\mathbf{A \, B \, f} = (\mathbf{B \, A} + [\mathbf{A,B}]) f$. The right hand side of that equation may be easier to solve, despite having more terms. Derivatives may also be written as operators, $\frac{d}{dx} f(x)$, and they also do not, in general, commute.
Sloppy notation about where an integrand (or the thing getting differentiated) ends creates ambiguity, so please put parentheses around the integrand if there is an addition or subtraction sign on that side of the equation.
In the integral you mentioned, the only reason to write it in operator notation is to introduce operator notation to you. I remember when they taught it to me, they were similarly (frustratingly) quiet about why they chose to write things that way. Bravo for asking the question.
It appears, in your edit, that you are asking about why the $dx$ 'commutes'. It may move around the integrand, but you should not interpret that as commutation. This is just a change in notation, and not in the intended meaning.
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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.
The meaning is the same.
Placing
dxat the beginning of the integral has advantages when you have nested integrals:$$\int_{x_0}^{x_1} dx \int_{y_0}^{y_1} dy \int_{z_0}^{z_1} dz \; f(x,y,z)$$
Otherwise, you need lots of parentheses in order to know which variable is integrated in which interval, and becomes less legible:
$$\int_{x_0}^{x_1} \left( \int_{y_0}^{y_1} \left( \int_{z_0}^{z_1} f(x,y,z) \; dz\right) dy\right)dx$$