Calc 2 fun (Length of a curve)

Ok here is my problem. I am in Calc 2 and I can not for anything figure out the math in this problem.

I have the forumal s=Int{a},{b}(Sqr(1 + f'(x)^2)dx
Note: Sqr = Squair root; Int = Integral

And I have f(x) = 1/4x^4 +1/8x^(-2) on [1,2]

So I can plug the function in and all but I can't figure out how to solve it. Please help I need to figure this out by class tomarrow. Note: This is not a problem that will be turned in don't feel like you are doing my homework for me. I just need help getting the concept of how to solve it.

I'm assuming you're in regular Cartesian system and integrating from a to b.

So first of all, you have the function f(x)
= 1/4 x^4 + 1/8 x^-2.
The length of curve formula requires the derivative of f(x)
f ' (x) = ( 1/4 x^4 )' + (1/8 x^-2 )'
= 4/4 x^3 + -2/8 x^-3
= x^3 - 1/4 x^-3

And the formula that you integrate is
Sqr(1 + f'(x)^2 )
= Sqr(1 + (x^3 - 1/4 x^-3)^2 )
= Sqr(1 + (x^6 - 1/4 x^3 x^-3 - 1/4 x^3 x^-3 + 1/16 x^-6) )
= Sqr(1 + (x^6 - 2/4 + 1/16 x^-6) )
= Sqr(x^6 + 2/4 + 1/16 x^-6)
= Sqr(x^6 + 1/2 + 1/16 x^-6)

Using my supreme mathematical skills, I spot that
x^6 + 1/2 + 1/16 x^-6
is equal to
(x^6 + 1/4)^2 / x^6

(I used the quadratic formula, actually, to get the 1/4, the x^6 terms due to there being an x^6 and a x^-6)

=Sqr[(x^6 + 1/4)^2 / x^6]
= Sqr[(x^6 + 1/4)^2] / Sqr[x^6]
= (x^6 + 1/4) / x^3
= x^3 + 1/4 x^-3

And then you integrate...
Integral from a to b w/ respect to x of x^3 + 1/4 x^-3

= 1/4 x^4 + 1/4 (-1/2) x^-2 (eval x from b to a)
= 1/4 x^4 - 1/8 x^-2 (eval x from b to a)
= 1/4 b^4 - 1/8 b^-2 - 1/4 a^4 + 1/8 a^-2
= 1/8 (2 b^4 - b^-2 - 2 a^4 + a^-2)
:)