The code below runs as expected up through the 4th line. At that point ans_all should contain all the multipule ansers to the equation being looked at. I want to look at just the first solution and I would expect the way to do this would be ans_all[1]. However the command ans_all[999999999] is valid, which indicates to me that that's not how you do that. So.... how do I get just the first answer?

equ := sqrt(1 + (-4 + 4*sqrt(2))*x + (16 - 12*sqrt(2))*x^2 + (-24 + 16*sqrt(2))*x^3 + (15 - 8*sqrt(2))*x^4)/(1 + (-4 + 4*sqrt(2))*x + (22 - 12*sqrt(2))*x^2 + (-36 + 28*sqrt(2))*x^3 + (33 - 20*sqrt(2))*x^4):
anit_dev_equ := int(equ,x):
ans_root := eval(anit_dev_equ, [x = 1]) - eval(anit_dev_equ, [x = 0]):
ans_all := allvalues(ans_root):
ans_all[999999999]