g(x)=x^3+(1/12x) for 1<x<3
I can get as far as integral of sqrt(1+9x^4-(1/12x^4)) before running into substitution problems... what should I set U to be? I tried using x^4 but I keep getting screwed up somewhere.. Thanks in advance!
$\displaystyle \displaystyle g(x) = x^3 + \frac{1}{12x}$
$\displaystyle \displaystyle g'(x) = 3x^2 - \frac{1}{12x^2}
$
$\displaystyle \displaystyle [g'(x)]^2 = 9x^4 - \frac{1}{2} + \frac{1}{144x^4}$
$\displaystyle \displaystyle 1 + [g(x)]^2 = 9x^4 + \frac{1}{2} + \frac{1}{144x^4} = \left(3x^2 + \frac{1}{12x^2}\right)^2
$
take it from here?