Calculus of variations: the inside function has an integral

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It is known that if the functional $$J=\int_a^b L(x,f(x))dx \tag{1}$$ has an extremum, then the Euler equation $\frac{\partial{L}}{\partial{f(x)}}=0$ holds. My question is, for example, what if $$L(x,f(x))=f(x) \int_a^b xf(x)dx~~ ? \tag{2}$$

(1) It seems to me that $\int_a^b xf(x)dx$ is still a function w.r.t $x$ and $f(x)$ denoted by $L(x,f(x))$.

(2) What is the corresponding Euler equation? Specifically, how to calculate $$\frac{\partial{[\int_a^b xf(x)dx]}}{\partial{f(x)}}~~~? \tag{3}$$

My question is similar to a previous post Calculus of Variations: Contains an integral of my goal function, which can be concluded that: the goal function has an integral of $f(x)$.