Arc Length by Integration 2

Let $L$ be the length of the curve $y= f(x)$ from $x = 2$ to $x = 8.$ Using the arc length formula, we have $L = \int_{2}^{8} \sqrt{1 + \left({f}'(x)\right)^2}\ dx ,$ where ${f}'(x) = \sqrt{{x}^4-1} .$ Thus,

$\begin{aligned} L = \int_{2}^{8} \sqrt{1 + \left({f}'(x)\right)^2} \ dx &= \int_{2}^{8} \sqrt{1 + \left(\sqrt{ {x}^4 -1 }\right)^2} \ dx \\ &= \int_{2}^{8} \sqrt{1+ {x}^4-1} \ dx \\ &= \int_{2}^{8} \sqrt{{x}^4} \ dx \\ &= \int_{2}^{8} {x}^2 \ dx \\ &= \left[\frac{1}{3} {x}^3 \right]_{2}^{8} = \frac{1}{3}\cdot ({8}^3 - {2}^3) \\ &= \frac{504}{3} =168. \end{aligned}$