Finding an Extremal for a function.

$I(y) = displaystyle int_{x_0}^{x_1} dfrac{1 + y^2}{(y')^3} dx$ reduced Euler-Lagrange equationd for $I=int F(x,y,y') dx $ is given as $$F-y'frac{partial F}{partial y'}=C.$$ So here we have $$frac{1+y^2}{y'^3}+3y'frac{1+y^2}{y'^4}=C implies y'= D(1+y^2)^{1/3}.$$ $$implies int frac{dy}{(1+y^2)^{1/3}}=Dx+E.$$ $$implies y~_2F_1(1/2,1/3,3/2;y^2)=Dx+E.$$ Here $~_2F_1(a,b;c,z)$ is the Gauss hypergeometric function.