Finding the length of the arc for function $8y = x^4 +2x^{-2}$

Question

Find the length of the arc of curve $8y = x^4 +2x^{-2}$ from $x=1$ to $x=2$

I first isolated for $y$ and derived:

$$ f(x) = {1 \over 8 }x^{4} + {1 \over 4}x^{-2} $$

$$f'(x) = {1 \over 2} x^3 - {1 \over 2}x^{-3}$$

Then found the arc length:

$$L = \int ^2 _1 \sqrt{1+[f'(x)]^2} dx \\ = \int ^2 _1 \sqrt{1+({1\over 2}x^3 -{1 \over 2}x^{-3})^2} dx \\ = \int ^2 _1 \sqrt{{1 \over 4} x^6 + {1\over 4}x^{-6} +{1 \over 2}} dx$$

Let $u = {1 \over 4} x^6 + {1\over 4}x^{-6} +{1 \over 2}$

$du = {{3 \over 2} x^5 - {3\over 2}x^{-7} dx}$

However, I cannot seem to integrate this

How can I integrate this equation?

Answer 1

Hint. Note that $${1 \over 4} x^6 + {1\over 4}x^{-6} +{1 \over 2}=\frac{(x^3+x^{-3})^2}{4}.$$ so it remains to evaluate $$L=\frac{1}{2}\int ^2 _1(x^3+x^{-3})\,dx.$$

Answer 2

Hint

$$u = {1 \over 4} x^6 + {1\over 4}x^{-6} +{1 \over 2}= \left({1 \over 2} x^3 + {1\over 2}x^{-3} \right)^2$$

Answer 3

Hint: $$2+x^6+\frac{1}{x^6}=\left(x^3+\frac{1}{x^3}\right)^2$$