Second Variation of Arc Length

Question

I'd like to find the second variation of the functional $$J=\int_0^1\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx$$ with boundary conditions $y(0)=0$, $y(1)=b.$ As an intermediate, one would typically look at $$\tilde{J}=\int_0^1\sqrt{1+\left(\frac{dy_*}{dx}+\varepsilon\frac{dh}{dx} \right)^2}dx,$$ where $y_*(x)$ is the function such that the first variation of the functional is zero, $h(x)$ is arbitrary, and $\varepsilon>0$. Letting prime denote derivative with respect to $x,$ and expanding the integral using the binomial series I believe gives $$\tilde{J}=\int_0^1\sqrt{\left(1+y_*'^2\right)\left(1+\frac{2\varepsilon h'y_*'}{1+y_*'^2}+\frac{2\varepsilon^2 h'^2}{1+y_*'^2}\right)}dx$$ $$=\int_0^1\sqrt{1+y_*'^2}dx+\varepsilon\int_0^1\frac{h'y_*'}{\sqrt{1+y_*'^2}}dx+\varepsilon^2\int_0^1\frac{h'^2}{2\sqrt{1+y_*'^2}}dx+\mathcal{O}\left(\varepsilon^3\right).$$ I know how to handle the first and second terms with integration by parts and such in the definition of the Frechet derivative, but I'm not sure what to do with the $\mathcal{O}\left(\varepsilon^2\right)$ terms. I was hoping to find an $h^2$ factor, but IBP doesn't seem to be the trick. What should I do?