= Solution
{c}
For the graph of a differentiable function, the <arc length> is $L=\int\sqrt{1+(y')^2}\,dx$. Since $y'=2x$,
$$
L=\int_0^1\sqrt{1+4x^2}\,dx.
$$
Use the <hyperbolic substitution> $2x=\sinh t$: then $dx=\tfrac12\cosh t\,dt$ and $\sqrt{1+4x^2}=\cosh t$. The <hyperbolic functions> $\cosh^2t=(1+\cosh2t)/2$ yields
$$
L=\left[\frac{x}{2}\sqrt{1+4x^2}+\frac14\operatorname{arsinh}(2x)\right]_0^1
=\boxed{\frac{\sqrt5}{2}+\frac14\ln(2+\sqrt5).}
$$
Here $\operatorname{arsinh}t=\ln(t+\sqrt{1+t^2})$ is the <inverse hyperbolic sine>.
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