Functional derivative of a functional that depends on antiderivative It is well known how to calculate functional derivative if a functional depends of the function and it's derivatives (Euler-Lagrange rule): $\mathcal{L}=\int F(x,\dot{x},t)dt$. There is also straight-forward generalization for a functional that depend on higher order function derivatives.
From now on I will use $x$ instead of $t$ and will denote functions by small english letters (e.g. $f$, $\phi$. etc.).
In my problem, a functional depends on the function's antiderivative:
$$
\mathcal{L}=\int dx \int_{-1}^x f(x')dx'.
$$
How can I calculate ${\delta \mathcal{L}}/{\delta f}$ ? I tried to do it by definition:
$$
\int \frac{\delta \mathcal{L}}{\delta f} \phi(x) dx = \left[\frac{dF[f+\epsilon\phi]}{d\epsilon}\right]_{\epsilon=0}.
$$
However, after simplifying the right hand side I found out that the result cannot be factorized to $\int (...)\phi(x)dx$. 
Denoting $F=\int_{-1}^xf(x')dx'$, I was also trying to use the chain rule, but did not succeed. 
In my actual problem I need to minimize $\mathcal{L}=\int dx\int \exp(-f(x'))dx'$.
Thanks,
Mikhail
Edit: Sorry, I should have been more specific from the beginning. The actual problem is find the functional gradient w.r.t. $f(x)$ of the following functional:
$$
\mathcal{L} = \int_{-1}^1 dx \exp\left[\int_{-1}^x f(x')dx'\right]\beta(x). 
$$
Note, that the second integral is entirely inside of the exponent. $\beta(x)$ does not depend on f(x).
 A: *

*OP first considers a functional $S\equiv {\cal L}$ of the form
$$\begin{align}S[f]~=&~\int_a^b\!\mathrm{d}x \int_a^x\!\mathrm{d}x^{\prime} ~g\circ f(x^{\prime}) \cr
~=&~\iint_{[a,b]^2}\!\mathrm{d}x ~\mathrm{d}x^{\prime} 
~\theta(x-x^{\prime})~g\circ f(x^{\prime}) \cr
~=&~ (b-x^{\prime}) \int_a^b\!\mathrm{d}x^{\prime}~g\circ f(x^{\prime}),
\end{align}
\tag{1a}$$
where $\theta$ is the Heaviside step function and $g$ is a fixed function. (In OP's examples $g(y)=y$ and $g(y)=e^{-y}$ and $a=-1$.)
The functional/variational derivative is then
$$\forall x^{\prime}~\in~[a,b]:~~\frac{\delta S[f]}{\delta f(x^{\prime})}
~=~ (b-x^{\prime}) ~g^{\prime}\circ f(x^{\prime}) .\tag{1b}$$

*In an edit OP considers a functional $S\equiv {\cal L}$ of the form
$$S[f]~=~\int_a^b\!\mathrm{d}x  ~g(x,F(x)), \qquad 
F(x)~:=~\int_a^x\!\mathrm{d}x^{\prime}~f(x^{\prime})
~=~ \int_a^b\!\mathrm{d}x^{\prime}~\theta(x-x^{\prime})~f(x^{\prime}). 
\tag{2a}$$
Let $g_F=\frac{\partial g}{\partial F}$ denote the partial derivative wrt. the second argument of the $g$-function.
The functional/variational derivative is then
$$\forall x^{\prime}~\in~[a,b]:~~\frac{\delta S[f]}{\delta f(x^{\prime})}
~=~  \int_a^b\!\mathrm{d}x~\theta(x-x^{\prime})~g_F(x,F(x))
~=~  \int_{x^{\prime}}^b\!\mathrm{d}x~g_F(x,F(x)).\tag{2b}$$
A: I am not convinced that calculus of variations will help you find a minimiser $f$ for your problem, as no minimiser exists. Consider
$$ \mathcal{L}[f]=\int_{-1}^1 \int_{-1}^x \exp(-f(x'))dx'dx
$$ 
Clearly $\mathcal{L}[f]\ge 0$ for all functions $f$. However, choosing $f_n\equiv \ln n$ (constant function), we have 
$$\mathcal{L}[f_n]=\int_{-1}^1 \int_{-1}^x \frac1n dx' dx = \frac1n\int_{-1}^1 (x+1)dx = \frac2n, 
$$
which tends to $0$ for $n\to\infty$. So if there is a minimiser, the value of the minimum must be zero. Hence we must have 
$\int_{-1}^x \exp(-f(x'))dx'=0
$ for almost every $x$, which implies $\exp(-f(y))=0$ for almost every $y$. The exponential is never zero, so no such function exists.
