Answer
$$\dfrac{53}{6}$$
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 $$L=\int_{0}^{2} [\dfrac{(2x^2+2x+)^2 \cdot (2x^2+6x+5)^2}{16(x+1)^4}] dx $$
Now, we will separate the terms and then integrate:
$$L=\int_{0}^{2} x^2+2x+1+\dfrac{1}{4(x+1)^2} dx \\=[\dfrac{x^{3}}{3}+x^2+x-\dfrac{1}{4(x+1)}]_{0}^{2} \\=\dfrac{53}{6}$$
