Hmm... I did it by hand (subsitution, t = g + 1/2 kA/m v^2) and got this:
y = -\frac{m}{kA}ln(1 + \frac{kAD}{m})
Which is the same as mathematica returned, but with an extra minus sign. Is it an error on my behalf?
Oh, I see. That's odd. Are you sure that in the text they didn't mean that the result from the arclength formula should be x^2/2 + lnx/4 before you put in the limits?
I don't see the problem, I don't get an extra 1/2 at the end. I get the same result as you do finding the function but I don't have any problem inputing back into the equation.
f(x) = x^2 - \frac{ln|x|}{4}
\frac{df(x)}{dx} = x - \frac{1}{4x}
\frac{df(x)}{dx}^2 = x^2 - \frac{1}{2} +...