Calculus 8th Edition

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