Answer
$\frac{5}{3}$
Work Step by Step
Setup the integration to find the length of the arc on the interval [0,1].
$y'= (x^2 +1)^{\frac{1}{2}}(2x)$
$ \int_0^1 \sqrt {1+[2x(x^2+1)^{\frac{1}{2}}]^2} dx$
$ \int_0^1 \sqrt {1+4x^2(x^2+1)}dx$
$\int _0^1 \sqrt {1+4x^4 +4x^2}dx$
$2\int_0^1 \sqrt {x^4 +x^2 + \frac{1}{4}} dx$, Take out 2 out of the integrand
$2\int_0^1(x^2 +\frac{1}{2})dx$ , integrate
$2[\frac{1}{3}x^3 + \frac{1}{2}x]_0^1$
$2(\frac{1}{3} + \frac{1}{2})-0$
$\frac{5}{3}$
