Surface Area of a Revolution

Find the surface area given by rotating the following curve about the y-axis.

y=$\displaystyle \frac{x^4}{4} + \frac{1}{8x^2} $ $\displaystyle 0\leq x \leq 3$

$\displaystyle \frac{dy}{dx}=x^3-\frac{1}{4x^3}$

$\displaystyle (\frac{dy}{dx})^2=x^6-\frac{1}{2x^6}-\frac{1}{2}$

$\displaystyle s=\int^3_0 2\pi*x*\sqrt{1+x^6-\frac{1}{2x^6}-\frac{1}{2}}$

$\displaystyle s=\int^3_0 2\pi*x*\sqrt{x^6-\frac{1}{2x^6}+\frac{1}{2}}$

That is as far as I can get. Can someone tell me how to take this integral? I need to take it just using algebraic manipulation and u-substitution if necessary. Thanks