Answer
$$\dfrac{99}{8}$$
Work Step by Step
The arc length can be calculated as follows: $L=\int_{a}^{b} \sqrt {1+[f'(x)]^2} dx$
Re-arrange the given equation such that $[f'(x)]^2=\dfrac{(y-1)^2}{4y}$
After simplification, we have $$L=\int_{1}^{8} [1+\dfrac{(4x^{2/3}-1)^2}{16x^{2/3}}] dx $$
Now, we will separate the terms and then integrate:
$$L=\int_{1}^{8} \dfrac{4x^{2/3}+1}{4x^{1/3}} \space dy \\=[ \dfrac{3x^{(4/3)}}{4}-\dfrac{3x^{(2/3)}}{8} ]_1^{8} \\=\dfrac{99}{8}$$
