find the arc length of the curve y = lnx for x = 1 to x = square root of 8
y' = 1/x
ds^2 = 1+y'^2 = 1 + 1/x^2
s = ∫[1,√8] √(1 + 1/x^2) dx
wow - see what wolframalpha came up with:
http://www.wolframalpha.com/input/?i=%E2%88%AB[1%2C%E2%88%9A8]+%E2%88%9A%281+%2B+1%2Fx^2%29+dx
http://www.wolframalpha.com/input/?i=arc+length+ln%28x%29+for+x%3D1+to+%E2%88%9A8
This kind of thing usually calls for a trig substitution, such as
u = tan(x) or sinh(x)