Firstly, we studied that integration is always with $\mathrm{d}x$. But in physics, I have seen this thing
$$V(t) = \frac{\mathrm{d}x}{\mathrm{d}t}$$ $$\mathrm{d}x = V(t)\ \mathrm{d}t$$ Then here he adds the integration $$\int\mathrm{d}x = \int V(t)\ \mathrm{d}t$$ Here I did not understand how to enter $\int$ without $\mathrm{d}x$. Here I wonder what this is $\mathrm{d}x$,
is that true : $$\int\frac{\mathrm{d}x \ \mathrm{d}t}{\mathrm{d}t} =∫\mathrm{d}x = x \,?$$
Please help me about everything you mentioned.
To answer the question of your title, the reason we use $dx$ as part of the notation for the integral is because integral is related to sums of the areas of very thin vertical triangles of very small width. Such an area is height times width $dx$. Leibniz originally introduced the symbol $\int$, which is an elongated $S$ because $S$ is the initial letter of the word "sum" (or its equivalent in Latin or French). More precisely, we partition the interval of integration on the $x$-axis into subintervals of very small length $dx$ and then take the corresponding sum.
To answer your additional questions, there is of course nothing special about the letter "x". If we think of the line as the $t$-axis, we would denote the corresponding partition length $dt$, and write the integral as $\int_a^b f(t) dt$, instead.
To answer your question about replacing $dx$ by $dt$ in the integral, note that one can further subdivide the subintervals of length $dx$ into, say, $N$ smaller subintervals of length $dt$ so that $dx= N\,dt$ or $\frac{dx}{dt}=N$, and then the corresponding sum would involve $\frac{dx}{dt} dt$ in place of $dx$. This corresponds to the change of variables formula in integrals.
The mathematical formalisation of such "very small subintervals" involves infinitesimals; you may want to look up some questions under the tag infinitesimals.