Very few functions y = f(x) lead to integrands of the form V1 + (dy/dx) which have elementary antiderivatives. In problem 13 , 1 + (dy/dx) is a perfect square and the resulting arc length integrals can be evaluated using antiderivatives. Do so. 5 13. y = 5 + 3 for 1 s x 5 5. 12x