Can I bring the variable of integration inside the integral?
No; since: $$\int_0^1x^2 \,\mbox{d}x=\frac{1}{3}$$ you would have: $$x\int_0^1x^2\,\mbox{d}x=\frac{1}{3}x$$ while: $$\int_0^1x^3\,\mbox{d}x=\frac{1}{4}$$
Or to put it differently: if you would be allowed to bring in such a factor, you would also be allowed to take out any factor and it might be easier (?) to understand that this can't work...
Also note that in a definite integral, the variable is just a dummy so it would be less confusing to not use the same variable but rather write something like: $$x\int_0^1t^2\,\mbox{d}t=\frac{1}{3}x$$ Now, as AccidentalFourierTransform pointed out in a comment, you can bring the $x$ inside if $x$ and $t$ are independent since in that case, $x$ is a constant with respect to $t$ and the integral is linear: $$x\int_0^1t^2\,\mbox{d}t=\int_0^1xt^2\,\mbox{d}t=\frac{1}{3}x$$
It is better to understand more deeply about this whole "variable of integration" thing. The definite integral denoted by symbol $$\int_{a}^{b}f(x)\,dx\tag{1}$$ is actually dependent only on two things: the function $f$ being integrated and the interval $[a, b]$ of integration. The most common way to represent a function is by giving an explicit formula to compute the image under the function written using some variable $x$ and then $f(x)$ denotes the image of $x$ under $f$. Thus in equation $(1)$ the symbol $x$ in $f(x)$ and in $dx$ is dummy. It serves no other purpose than to represent the function. In fact the whole $dx$ does not serve any purpose at all except that it helps to memorize the technique of substitution in evaluation of integrals.
Thus we may well represent the integral in $(1)$ just via the symbol $$\int_{a}^{b}f\tag{2}$$ if we know the function $f$ being talked about. But as I said earlier most convenient notation for functions has turned out to be giving a formula for image under $f$ using some variable so that the equation $(1)$ is the preferred notation compared to $(2)$.
But when we use an independent variable just to express the function, we must understand that it does not serve any more purpose than that. The variable $x$ in your question which lies outside the integral serves a different purpose. Most probably it is representing some real number in a given context and the integral after it is just another real number. But if we bring the $x$ (lying outside the integral) inside the integral sign it does interfere with the expression of the function being integrated. It is best to use a different variable under the integral sign to avoid confusion.
The same problem can be seen in the case of limit operation when we are dealing with an expression like $$x\lim_{x \to a}f(x)\tag{3}$$ Again understand that the expression $\lim_{x \to a}f(x)$ is dependent only on function $f$ and point $a$ and the symbol for a variable $x$ is used only because that is the most convenient way to represent a function and the variable under limit operation is thus a dummy variable and if there is a need (to avoid confusion) we may well replace it with any other symbol we wish to use.
You can also give similar remarks about the notation $$\frac{d}{dx}f(x) = g(x)\tag{4}$$ and I leave it as an exercise for you.
I like to write integrals as operators on functions, as much as possible1:
$$\int_0^1{x^2}\,\mbox{d}x = \int_0^1f$$ with $$f(x) = x^2.$$
Then your question becomes whether
$$x\int_0^1f = \int_0^1{xf}.$$
Well... because there is no "integration variable", it is clear that $x$ is a constant for the purposes of integration. So, written like that, it is ok. It is the same as $\int_0^1xt^2\,\mbox{d}t$ in StackTD's answer.
On the other hand, the $\mbox{d}x$ inside the integral "captures" your $x$, and it would be interpreted as being another appearance of the integration variable. The meaning of the expression changes.
1. At least for single-parameter functions...