Help me understand the derivation of the kinetic energy formula please In my physics textbook, kinetic energy is defined as $W$Net $=$ $\int m\frac {dv}{dt}$ $dx$ This makes sense to me just fine. The book goes on to rearrange the integral to say the following:
$W$Net $=$ $\int m\frac {dx}{dt}$ $dv$  And since $\frac {dx}{dt} = v$, the integral becomes $W$Net $=$ $\int mv$ $dv$
I've always been taught that the notation $\frac {d}{dx}$ is not a fraction and shouldn't be treated as one, however the top answer on  this post helped me understand that it's simply an application of the chain rule. However, the last two steps have me confused:
$W$Net $=$ $m\int v\frac {dv}{dx}$ $dx$ 
And then the answer says "the $dx$ cancels, leaving you with
$m\int vdv$

However, as I understand it, the $dx$ can't just "cancel", and furthermore, shouldn't the integral of
$\int v\frac {dv}{dx}$ $dx$ when integrating over $dx$ be equal to $(\frac{v^2}{2})(v)$, since the integral of $\frac{dv}{dx}$ is just $v$ and the integral of $v$ should be $\frac{v^2}{2}$?
What is the flaw in my logic? I'm sick of textbooks and tutors saying things like "dx cancels" when it doesn't.
 A: As pointed out in the answer by stackexchange contributor Maximal Ideal, in the steps you are concerned about the main violation of proper mathematical procedure is that the integration is declared without start point and end point. A textbook author who does that paints himself into a corner.
Her is an instance of the derivation with limits applied from the start, and carried throughout:

The starting point for deriving the Work-Energy theorem is Newton's second law:
$$ F = ma \tag{1} $$
Integrate both sides with respect to the spatial coordinate, integrating from starting point $s_0$ to final point $s$
$$ \int_{s_0}^s F \ ds = \int_{s_0}^s ma \ ds \tag{2} $$
At this point we can develop the right hand side, by capitalizing on the fact that position and acceleration are not independent of each other.
$$ v = \frac{ds}{dt} \quad \leftrightarrow \quad ds = v \ dt  \tag{3}  $$
$$ a = \frac{dv}{dt} \quad \leftrightarrow \quad  dv = a \ dt  \tag{4}  $$
In the following the relations (3) and (4) are used for substitution. That is: the steps are instances of integration by substitution; the simplest possible instance of integration by substitution.
I omit the factor $m$ temporarily, it is a multiplicative factor that is just carried over each step
$$ \int_{s_0}^s a \ ds \tag{5}  $$
Use (3) to change the differential from $ds$ to $dt$. Since the differential is changed the limits change accordingly.
$$ \int_{t_0}^t a \ v \ dt \tag{6} $$
Change the order:
$$ \int_{t_0}^t v \ a \ dt  \tag{7} $$
Change of differential according to (4), with corresponding change of limits.
$$ \int_{v_0}^v v \ dv  \tag{8} $$
So we have:
$$ \int_{s_0}^s a \ ds = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2  \tag{9} $$
We multiply both sides with $m$, and then the right hand side of (9) gives us the right hand side of (2). The result: the Work-Energy theorem:
$$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2  \tag{10} $$
A: $\frac{d}{dx}$ is a limit and not fraction. However, if you have been introduced to limits and calculus, you have been surely told about conditions in which limits can be distributed. Due to this, the $\frac{d}{dx}$ looks very similar to fractions and you can safely do this in this case. It's worth looking at Math's SE post   here.
"shouldn't the integral of
$\int v\frac {dv}{dx}$ $dx$ when integrating over $dx$ be equal to $(\frac{v^2}{2})(v)$, since the integral of $\frac{dv}{dx}$ is just $v$ and the integral of $v$ should be $\frac{v^2}{2}$?"
I am not sure what you are trying to say. I am afraid that you may not be aware of integration techniques yet.
"like "dx cancels" when it doesn't."
Yes , they don't just cancel. In fact, what is going on is that : $\frac{d}{dx}$($x$) = $1$
A: Let me first pick up where you were left (rightfully) confused.

$W$Net $=$ $m\int v\frac {dv}{dx}$ $dx$ 
And then the answer says "the $dx$ cancels, leaving you with
$m\int vdv$

However, as I understand it, the $dx$ can't just "cancel" [...]

Yes, you are completely correct in saying that you can't just blindly say "it cancels" (as many people and textbooks might do). Since we're working with an integral, the correct way to proceed is to consider substitution of variables.
We have the integral
$$ W_{\text{Net}} = m\int_{x_{i}}^{x_{f}} v\frac{dv}{dx} \, dx $$
where I decided to include the definite limits of integration. Here $x_{i}$ is the initial position and $x_{f}$ is the final position.
Evidently the variable $v$ is meant to be taken as a function of $x$ in the integrand, so we'll write it out as
$$ W_{\text{Net}} = m\int_{x_{i}}^{x_{f}} v(x)v'(x) \, dx. $$
Now look at the wikipedia article for how substitution is properly done: https://en.wikipedia.org/wiki/Integration_by_substitution#Definite_integrals

$$ \int_a^b f(\varphi(x))\varphi'(x)\, dx = \int_{\varphi(a)}^{\varphi(b)} f(u)\,du. $$

We can match our integral to the LHS integral by taking $f(w) = w$ (so $f$ is just the identity function) and $\varphi(x) = v(x)$. Then
$$ W_{\text{Net}} = m\int_{x_{i}}^{x_{f}} v(x)v'(x) \, dx = m\int_{v(x_{i})}^{v(x_{f})} u \, du. $$
This last integral can be evaluated directly and I recommend you do it yourself. You should immediately obtain the work-energy theorem.
However, if you wish, we can do some "notation cleaning" to match what is written in physics texts. First, we know $v(x_{i})$ is the velocity at initial position of the path, so it is just the initial velocity — we'll label it as $v_{i} = v(x_{i})$. Similarly we label the final velocity as $v_{f} = v(x_{f})$. Inside the integrand, the letter being used doesn't matter, but since the limits of the integral are velocities, we know that the meaning of the variable being used is velocity (to be more specific $du$ must have units of velocity). Thus, it makes sense to relabel $u$ as the letter $v$ to carry the connotation of velocity. From this we get
$$ W_{\text{Net}} = m\int_{v_{i}}^{v_{f}} v \, dv = m \left( \frac{v_{f}^{2}}{2} - \frac{v_{i}^{2}}{2} \right) = \frac{1}{2}mv_{f}^{2} - \frac{1}{2}mv_{i}^{2}, $$
which is the work-energy theorem, as desired.

Now let me address another point that you raised.

[...] and furthermore, shouldn't the integral of
$\int v\frac {dv}{dx}$ $dx$ when integrating over $dx$ be equal to $(\frac{v^2}{2})(v)$, since the integral of $\frac{dv}{dx}$ is just $v$ and the integral of $v$ should be $\frac{v^2}{2}$?

No, this is wrong. When you have a product inside an integral, you cannot integrate each factor individually. Explicitly,
$$ \int_{a}^{b} f(x)g(f) dx \ne F(x)G(x)|_{a}^{b}. $$
Instead, the usual right way to handle products is to use integration by parts. Now it is not always needed, and in our case we definitely don't need it.

What I hope you understand is that this talk of "canceling $dx$ in the integral" is a actually shorthand for u-substitution just like how "treating derivatives like fractions" is a shorthand for the chain rule.
As you can see, it's a bit more complicated when you do all this rigorously.
I applaud your insistence on rigor here, and I hope you continue to probe these details. It will further your understanding. However, at the same time I should also say that there is value in these shorthands, and you should know how to work with them as well. Otherwise, you will end up far too inflexible.
True understanding comes from being comfortable with both approaches (the rigorous way and the handwavy shorthand way) and knowing how to synthesize them together.
A: "I've always been taught that the notation
$ \frac{d}{dx} $
is not a fraction and shouldn't be treated as one"
In physics class, it's works well in 1 dimension.
In practice? yes
In rigorious math class? no
