Does $\int e^{dx}$ makes sense? I know this is quite weird or it does not make much sense, but I was wondering, does $\int e^{dx}$ has any meaning or whether it makes sense at all? If it does means something, can it be integrated and what is the result?
 A: The differential in an integral is essentially a symbolic way to show on which variable you integrate and is not to be taken as a factor. With this in mind,
$$\int e^{dx}$$ is just syntactically unparsable, in the same way as
$$\sin(x=(+$$
A: Well it makes sense when you think about it one way and it doesn't when you think about it another way...First of all $dx$ is supposed to be extremely close to $0$ so (as J.G. said)
that's going to equal $1+dx$ which is essentially 1. On the other hand its not supposed to make sense cuz $e^{dx}$ it self doesn't have a value (you can't take it as $1$ because $dx$ is not $0$ and you can't take it as a value because $dx$ is super small) so how can you actually add up infinite amounts of this...I mean integration is essentially zooming in so that the area under a graph is actually infinite amounts of small rectangles but with the $dx$ difference you can't zoom in infinitely because there will always be the difference...I mean you can zoom in on $e^2$ because it has a fixed value...but you can't do that for $e^{dx}$ because $dx$ doesn't really have a well defined value
Hope you got your answer
A: I agree, that formally it make no sense, but as lunch exercise in mathematical fantasy , if we can give some sense to $(dx)^n$ as some measure, then we can imagine $\int e^{dx}=\int \sum \frac{(dx)^n}{n!}$. Now the point is what is $(dx)^n$..
A: Tool 1: Know, $\displaystyle \lim_{\Delta x\rightarrow 0} \frac{a^{\Delta x}-1}{\Delta x}=\ln(a) $. Now, \begin{align*} 
\int e^{\mathrm{d}x}&=\int e^{\mathrm{d}x}-1+1\\
&=\int \frac{e^{\mathrm{d}x}-1}{\mathrm{d}x}\mathrm{d}x+1\\
&=\int \ln(e)\,\mathrm{d}x+1\\
&=x\ln(e)+1\\
&=x+1 +\text{constant}
\end{align*}
