x'=\dfrac{1}{3}\cdot \dfrac{3}{2}(y^{2}+2)^{\frac{1}{2}}\cdot 2y = y\sqrt{\,y^{2}+2\,}.

1+(x')^{2} = 1 + y^{2}(y^{2}+2) = y^{4} + 2y^{2} + 1 = (y^{2}+1)^{2}

L=\displaystyle \int_{1}^{2}\sqrt{(y^{2}+1)^{2}}\,dy = \int_{1}^{2}(y^{2}+1)\,dy = \left[ \dfrac{y^{3}}{3} + y \right]_{1}^{2}

L=\left(\dfrac{8}{3}+2\right) - \left(\dfrac{1}{3}+1\right) = \dfrac{7}{3}+1 = \dfrac{10}{3}