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I need to minimize integral: $$\int_{0}^{1} F(x,u,u') dx=\int_{0}^{1}(u^2+2xu'u)dx \to\inf$$ with the boundary conditions $$u(0)=0, u(1)=1$$ Algorithm is to write solve the Euler–Lagrange equation, then get the correct constant with respect to boundary conditions, however: $$\int_{0}^{1}(F_uh+F_u'h')dx=\int_{0}^{1}((2u+2xu')h+2(xu)h')dx=2\int_{0}^{1}((xu)'h+(xu)h')dx=\int_{0}^{1}(xuh)'dx=1*u(1)h(1)-0=u(1)h(1)$$ ,but we require $h(0)=h(1)=0$, so Euler–Lagrange equation identically equal to $0$

What should i do next?

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    The lagrangian is the total derivative respect $x$ of $xu^2$ Two lagrangians give the same equations if they differ in a total derivative respect $x$ of a function of the variable and the function to minimize. The simplest lagrangian is $0$ in this case. – 2017-02-20

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$$ \int_0^1 2 x u' u dx = \int_0^1 x d(u^2)=\left. x u^2 \vphantom{\int}\right|_0^1 -\int_0^1 u^2\,dx=1-\int_0^1 u^2\,dx. $$ This means that $$ J[u]=1 $$ for all $u \in C^1([0,1])$ satisfying the boundary conditions. So the functional is constant on the class of the admissible functions. Your argument shows that the Gateaux/Frechet variation vanishes identically on the set of admissible functions, no wonder here.