Isoperimetric problem with auxillary conditions

Find the extremals for the problem with the functional $$I[x] = \int^1_0 [\dot{x}^2 - x^2 ] dt$$ and auxillary and boundary conditions $$\int^1_0 \sqrt{1+\dot{x}^2} dt = \sqrt{2}\hspace{5pt} ,\hspace{5pt} x(0)=0 \hspace{5pt} ,\hspace{5pt}x(1) = 1$$ What I have done: Using $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right) - \frac{\partial L}{\partial x} - \lambda \left[\frac{d}{dt}\left(\frac{\partial G}{\partial \dot{x}}\right) -\frac{\partial G}{\partial x} \right]=0$$ where $L$ is the functional and $G$ the auxillary condition, I get $$2\ddot{x}+2x - \lambda \left[ \frac{d}{dt}\left(\frac{\dot{x}}{\sqrt{1+\dot{x}^2}} \right) \right] =0$$ Taking the derivative $$2\ddot{x}+2x - \lambda \left[ \ddot{x}(1+\dot{x})^{-\frac{1}{2}} - \dot{x}^2\ddot{x}(1+\dot{x})^{-\frac{3}{2}} \right] =0$$ I don't know how to simplify this to be able to integrate. The best I could do was multiply through by $(1+\dot{x})^{\frac{3}{2}}$ giving $$(2\ddot{x} + 2x)(1+\dot{x}^2)^{\frac{3}{2}} - \lambda\ddot{x}=0$$ but all my attempts to simplify it have lead to it being way more complicated.