Answer
$\sqrt{2} + \ln({1+\sqrt{2}})$
Work Step by Step
$y = \frac{2}{1} x^{2}$
then $y' = x$ and $1+(dy/dx)^{2} = 1 + x^{2}$
So $L = \int^{1}_{-1} \sqrt{1+x^{2}} dx = 2 \int^{1}_{0} \sqrt{1+x^{2}}dx$
$ = 2[\frac{x}{2} \sqrt{1+x^{2}} + \frac{1}{2} \ln(x+\sqrt{1+x^{2}})]^{1}_{0}$
$= \sqrt{2} + \ln({1+\sqrt{2}})$
