What is the length of the curve, y=1/3(x^2+2)^(3/2), from x=0 to x=9?

Answer 1

y = 1/3 (x^2+2)^(3/2)
y' = x(x^2+2)^(1/2) -- how convenient!

ds = (1+y'^2)^(1/2)
= (1+x^2(x^2+2))^(1/2)
= (x^2 + 1)

now it's smooth sailing