I was reading this https://math.iupui.edu/~jwatt/16500/honors/HP13.pdf and stuck on page 5 equation$(2)$: $$ E = m_0 \int_0^{x} \frac{v_1}{\left(1- \frac{v_1^2}{c^2} \right)^{3/2}}\, \frac{dv_1}{dx}\, dx = m_0 \int_0^{v_1} \frac{v_1}{\left(1- \frac{v_1^2}{c^2} \right)^{3/2}}\, dv_1 $$ It said using chain rule, it convert the integration from 0 to $x$ to 0 to $v_1$?
As my understanding "$x$" suppose to be distance while "$v_1$" suppose to mean velocity
Well, this is just a classical $u, du$ integration (or integration by substitution). Take $u(x) = v_1(x)$ with $v_1$ as a function of $x$ (see the last part of my answer). Thus, $du = \frac{dv_1}{dx}dx$ (by the chain rule), and then you obtain such equation. Note that when performing a change of variable like this (substitution), the integral limits will change accordingly.
EDIT: Let me add more detail of this procedure, in the sense of the examples of the wikipedia link I provided. Recall that we chose $u = v_1(x)$. Hence, we transform the integral as: $$ \int_{0}^x \frac{v_1(x)}{\left(1-v_1(x)^2/c^2\right)^{3/2}}\frac{dv_1(x)}{dx}dx = \int_{u=v_1(0)}^{u=v_1(x)} \frac{u}{\left(1-u^2/c^2\right)^{3/2}}du $$ Now, as stated in the document you provided, the object starts at rest. This means that the initial velocity $v_1(0)=0$. Hence, we simply obtain: $$ \int_{0}^{v_1} \frac{u}{\left(1-u^2/c^2\right)^{3/2}}du $$ with $v_1(x)=v_1$ for short. You may be confused (or not) regarding that this last integral doesn't look like the one one your question, since it has $u$ instead of $v_1$. But you can just substitute $u\to v_1$, since the "name" of the variable in an integral (the name of the variable for which you are integrating) is not important (its a "free" variable). See for example this question. As a remark, note that you can change variable names when you are integrating over them, but you can't change them in the limits of the integral for example. Take for instance the first expression of the integral, and make the change $x\to x'$: $$ \int_{0}^x \frac{v_1(x)}{\left(1-v_1(x)^2/c^2\right)^{3/2}}\frac{dv_1(x)}{dx}dx=\int_{0}^x \frac{v_1(x')}{\left(1-v_1(x')^2/c^2\right)^{3/2}}\frac{dv_1(x')}{dx'}dx' $$ Note that we changed $x$ to $x'$ in all the expression except for the $x$ in the integral limits. In other words, the $"x"$ inside the integral, is not the same $"x"$ in the limits, which might be the origin of the confusion.
Regarding to the last part of the question. I don't think $v_1$ is mean velocity. Or maybe I don't understand what do you mean by mean velocity then. In the same page of the document you mentioned, the author states that he is considering an object with mass $m_1$ with position $x$ and (instantaneous) velocity $v_1$. This is, $v_1$ is by nature a function of time. However, since $x$ is also a function of time you can write $$ a_1 = \frac{dv_1}{dt} = \frac{dv_1}{dx}\frac{dx}{dt} = v_1\frac{dv_1}{dx} $$ since $v_1 = \frac{dx}{dt}$ (which you might take as a definition of $v_1$) and using the chain rule. Thus, you can think of $v_1$ as a function of $x$ instead of $t$. Let me know if this is useful to you.