The functional is:
$$L=\int{u\sqrt{1+(u')^2}}dx$$
when: $$u'(0)=0, u(1)=2 $$
the steps that I did: $$ F=u\sqrt{1+(u')^2} $$ $$ F_u=\sqrt{1+(u')^2} $$ $$ F_{u'}=\frac{uu'}{\sqrt{1+(u')^2}} $$
because $$F_x =0$$ I can use Beltrami identity instead of EL $$F-u'F_{u'}=c$$
$$u\sqrt{1+(u')^2}-u'\frac{uu'}{\sqrt{1+(u')^2}}=c$$
$$\frac{u}{\sqrt{1+(u')^2}}=c$$
$$\frac{u}{\sqrt{1+(u')^2}}=c$$
$$\frac{u^2}{c^2}=1+(u')^2$$
$$u'=\pm\frac{\sqrt{u^2-c^2}}{c^2}$$
$$\frac{du}{dx}=\pm\frac{\sqrt{u^2-c^2}}{c}$$
$$dx=\pm c\frac{du}{\sqrt{u^2-c^2}}$$
$$x=\pm c\int{\frac{du}{\sqrt{u^2-c^2}}}=\pm c \ln{|u+\sqrt{u^2-c^2}|}$$
in this step I can use $u(1)=2$ to find $c$, but I'm still not sure how to find the critical function.
but I feel that something wrong with this solution, any help welcome
Hint.
You need two integration constants.
$$ \frac{u}{\sqrt{1+u'^2}} = C_1 $$
and after
$$ dx = \pm \frac{C_1 du}{\sqrt{u^2-C_1^2}} $$
giving
$$ x = C_2 \pm C_1\ln\left|u+\sqrt{u^2-C_1^2}\right| $$
After inversion
$$ \cases{ u = \frac 12 e^{-\frac{C_2+x}{C_1}}\left(C_1^2 e^{\frac{2C_2}{C_1}}+e^{\frac{2x}{C_1}}\right)\\ u = \cdots } $$
and finally determine $C_1, C_2$ according to the boundary conditions.