Cancelling out constants with u-substitution integration When using u-substitution to integrate, I tend to think about adding constants to make my $dx$ match my $du$. I don't have a basic enough understanding to grasp why it won't work if my $du$ contains the variable I am integrating with respect to.
My intuition tells me that adding to the function the variable that I am integrating is changing the function itself (meaning I have fundamentally modified the nature of the function, rather than adding constants which would just scale it).
For example:
$$\int e^{x^2} dx $$
$$u = x^2, du = 2xdx $$
Why can I not just say $dx = du/2x$?
 A: This is just a personal opinion.
I must confess that, when I started working on Mathematics Stack Exchange, I have been  surprised to see how the "u" substitution was used (and then taught).
When I was young (that is to say long time ago !), the way we were taught was quite different. It was like that
$$u=f(x) \implies x=f^{(-1)}(u)\implies dx=\frac{du}{f'\left(f^{(-1)}(u)\right)}$$
For example, using the case you give
$$u=x^2 \implies x=\sqrt u\implies dx=\frac{du}{2 \sqrt{u}}$$
Another example
$$u=\sin(x)\implies x=\sin ^{-1}(u)\implies dx=\frac{du}{\sqrt{1-u^2}}$$
For sure, this can make some calculations longer but I still think that it is clearer not to say more "natural".
A: The example is inappropriate as this does not have an elementary solution. I think you actually mean $\int e^{x^2} x dx$, substitution allows you see $1/2 \int e^u du$ then you can use the established formula to obtain the elementary solution.
There is nothing wrong saying $dx = du/2x$, it is just equivalent. $\int e^{x^2} x dx = \int e^{x^2} x \frac{1}{2x}  du = 1/2 \int e^u du$, you just cannot move the x out of integration as x is a variable.
A: We can start by making the substitution in question:
$$u = x^2, \,du=2x \,dx$$
Then
$$\int e^{x^2}\,dx =\int \frac{e^u}{2x} \,du$$
One might think this is easily integrable from here because we have only an $e^u$ and $du$, but we have forgotten something crucial: $x$ is not a constant; in fact,  $x = \sqrt{u}$. So
$$\int \frac{e^u}{2x} \,du = \int \frac{e^u}{2\sqrt{u}} \,du$$
A good rule of thumb for integrals is that if the variable changes as $u$ changes, then you cannot ignore it. Also, if you're curious, the integral in question cannot be expressed in a closed form by elementary functions.
