I watched this video which showed how one could evaluate $\int x^{dx}-1$:
$$\int x^{dx}-1$$ $$=\int \left(x^{dx}-1\right)\frac{dx}{dx}$$ $$=\int \frac{x^{dx}-1}{dx}dx$$
By Riemann Integration we have
$$\int^b_a f(x)dx=\lim_{n\to\infty}\sum_{i=1}^{n}f(x_i)\Delta x$$
Where $x_i=a+i\Delta x$ and $\Delta x=\frac{b-a}{n}$. Thus $\lim_{n\to\infty}\Delta x=dx$. Therefore, we can rewrite the integral as
$$\int \lim_{\Delta x\to0}\frac{x^{\Delta x}-1}{\Delta x}dx$$
(Note that $\lim_{\Delta x\to0}\frac{x^{\Delta x}-1}{\Delta x}=\ln x$)
$$=\int \ln xdx$$ $$=x\ln x-x+C$$ I am wondering about what would happen if instead of integrating $x^{dx}-1$ we integrated $f(x,dx)$ for some continuous function $f$. Using the same steps it seems like we would find that $\int f(x,dx)=\int \lim_{\Delta x\to0}\frac{f(x,\Delta x)}{\Delta x}dx$. However, the proof is not very rigorous, as dividing by $dx$ (an infinitesimal) is not valid.
Therefore, my question is: Is it valid to say that
$$\int f(x,dx)=\int \lim_{\Delta x\to0}\frac{f(x,\Delta x)}{\Delta x}dx$$
and what would be a rigorous proof?