Is the following statement true? $\int_{-1}^{x}\ {2x+x^2}\,dx = \int_{-1}^{x}\ F'(x) = F(x) = \int_{-1}^{x}\ {2t+t^2}\,dt$ I have trouble understanding easy integral concept.
Here, one of the integral theorem states:

And I was wondering what would happen if $F'(x)$ is back in to the integral.
For example, $F(x) = \int_{-1}^{x}\ {2t+t^2}, x \in [-1,5]$
$$F'(x) = 2x + x^2, x \in (-1,5)$$
Then if $F'(x)$ is back in to $\int_{-1}^{x}$ as in $\int_{-1}^{x}\ {2x+x^2}$, should it return $F(x)$?
I mean is the following statement true? 
$$\int_{-1}^{x}\ {2x+x^2} = \int_{-1}^{x}\ F'(x) = F(x) = \int_{-1}^{x}\ {2t+t^2}$$
What I am confused about is the change of the variable above.
 A: 
I mean is the following statement true? $$\int_{-1}^{x}\ {2x+x^2}\,dx = \int_{-1}^{x}\ F'(x) = F(x) = \int_{-1}^{x}\ {2t+t^2}\,dt$$

Not quite: $\text{If}\; F(x) = \int_{-1}^x \,2t + t^2 \,dt$, while it is true that $F'(x) = 2x + x^2$, we cannot write, meaningfully, $$F(x) = \int_{-1}^\color{blue}{\bf x} 2\color{blue}{\bf x} + \color{blue}{\bf x}^2\,d\color{blue}{\bf x}$$
The reason $t$ is used in the original integral $F(x) = F(x) = \int_{-1}^{x}\ {2t+t^2}\,dt$ is merely to stand in as a "dummy variable" which is needed only until we compute the indefinite integral, evaluate at the bounds, which thus gives us a function of $x$: $F(x)$. $F'(x)$ is also a function of $x$. 
What you can say is that $F(x) + C = \int F'(x) \,dx$, since here we are not using bounds of integration that are nonsensical.
We can integrate the derivative of a function, and we can find the derivative of the integral of a function, yes indeed. One can think of them, informally, as inverse processes given a function like the one you've posted.
A: There's at least one problem with this: you have an expression of the form $$\int_c^x F'(x) \, dx,$$ and the variable occurs as the variable of integration and in the limits, which does not make sense. So if you change the name of one of the variables here, the question makes more sense.
But the answer is yes: $$\int_c^\xi \left( \frac{d}{dx} \int_c^xf(t)\,dt \right) \, dx= \int_c^\xi f(x) \, dx.$$
Clarification: The issue with the variables doesn't have to do with properties of the derivative at all. I am told that in the subject of mathematical logic, the issue as described as "free variable vs. bound variable". If you are evaluating $$\psi(\xi) = \int_c^\xi \varphi (x) dx,$$ it does not make sense call $\xi$ by the same name as $x$, because for each $\xi$ that you plug in, you need $x$ to vary over the interval of integration. It doesn't make sense to "plug in" a particular $x$ into the function $\psi$.
