Confusion and clarification: Calculus of variations - Stationary Points I'm slightly confused as to what i need to do in this question, i'm currently studying the Euler-Lagrange equation as part of Calculus of variations, i feel like i may be over thinking this problem so some clarification would be amazing. thanks! i've been asked the following: 

Find the stationary points of the functional $$J(u) = \int_{a}^{b}\left[ \frac{|u'|^{2}}{2} + uu'+u'+u\right] dx$$
  where the values of u at the endpoints are not specified.

Def: We say that $u \in V$ is a stationary point of the differentiable functional $$J:(V,\|\cdot\|) \longrightarrow \mathbb{R}$$ if $\delta J(V,\cdot)$ is the zero function, 
note: $\delta J(u,v)$ is the directional derivative of u in the admissible direction $v\in V$ and This essentially translates to the fact that a stationary function u solves the euler-legrange equation.
so, in this case we take the lagrangian $$\Lambda(x,u,u') = \frac{|u'|^{2}}{2} + uu'+u'+u $$
this gives:
$$\frac{\partial \Lambda}{\partial u} = u'+1$$ $$\frac{\partial \Lambda}{\partial u'} = u'+u+1$$
and
$$\frac{d}{dx}\frac{\partial \Lambda}{\partial u'} = u''+u'$$
then the EL equation is:
$$\frac{d}{dx}\frac{\partial \Lambda}{\partial u'} - \frac{\partial \Lambda}{\partial u} = u''+u' - u' - 1 = u'' - 1 = 0$$
so we have:
$$u(x) = \frac{x^{2}}{2} + c_1 x + c_2$$
so, from my understanding this equation is the stationary function which makes J the shortest/largest/point of inflexion in terms of "distance".  either way, this is the equation which solves the above definition. and so is the stationary point of the functional J.
now, my question is, does this solve the problem? finding the general solution of J and if we have aux conditions we can narrow it down further. the values of u arent specified so i have to leave it as is? or, at this point should i also be looking for the extremal values of u itself, which given its a quadratic would be a minimum.
Cheers for the help, i appreciate it.
 A: Need to be just on with the basic method. 
When $\Lambda $ does not involve $x$  explicitly, (Beltrami) integration of Euler-Lagrange is convenient:
$$ \Lambda - u'\frac{\partial \Lambda}{du'}=c_1 $$
$$\frac {u^{'2}}{2}+ u u'+u'+u -u'(u'+u+1) = c_1$$
$$u'= \sqrt{2 (u-c_1)}$$
$$ \int \frac{du}{\sqrt{u-c_1}} = \sqrt{2}x  $$
EDIT1:
$$ \sqrt{u-c_1}=\frac{x}{\sqrt 2}+c_2$$
Squaring to simplify.. and absorbing arbitrary constants
$$ u=\frac {x^{2}}{2}+ c_3x + c_4$$
exactly same as what you have obtained except arbitrary constants.
A: FWIW, note that boundary conditions (BCs) are important.
If there are no BCs, then OP's functional $J[u]$ becomes just $(b-a)u$ for a constant path $u$. If $a\neq b$, this is unbounded both from above and below. For similar reasons there are then no stationary paths.
So to have stationary paths, one must impose BCs. So when the problem formulation says that the values of $u$ at the endpoints are not specified, it must mean that the boundary values of $u$ are fixed, but that the fixed values are unknown.
BCs are also needed in the proof of the Euler-Lagrange (EL) equation, or Beltrami's identity. See Narasimham's answer for details.
