Calculus arc length , can't find anti-derivitive?

Question

Find the arc length of the curve
f(x)=x^4 /8 + 1/(4x^2)
over the interval [2,3]

Using the arc length formula  ∫√(1+ f'(x)^2)  
I get this as the integrand after differentiating the function, squaring it, and simplifying everything under the radical.

∫√(1/2 + 1/4 x^6 - 1/(4x^6))

I cannot find the anti-derivative of this.   The method should be fairly simple since this problem assumes no more than basic substitution or numerical integration.

MechEng2030 · Accepted Answer

Okay, so what you have so far is almost right except it should be 1/(4x^6) at the end since (-1/2x^3)^2 = 1/(4x^6). Good work nonetheless. Now you just need some simplification to make this integral a little less complicated. I suggest you get a common denominator of 4x^6. After doing that, you get:

∫√((2x^6 + x^12 + 1)/(4x^6)) dx

Now take the square root of the numerator and denominator to get:

∫√(x^12 + 2x^6 + 1)/(2x^3) dx

Now note that x^12 + 2x^6 + 1 = (x^6 + 1)^2, so:

∫(x^6 + 1)/2x^3) dx from 2 to 3

= ∫(1/2 * (x^3) + 1/2 * x^-3)) dx

= 1/2 * ((x^4)/4 - 1/2 * x^-2 eval. from 2 to 3))

= 1/2 * (81/4 - 1/18 - 4 + 1/8))

= 1175/144

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Calculus arc length , can't find anti-derivitive?

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