Past exam of the mathematics course of the University of Oxford / Undated / Prelims / Mathematics III resit / 2 / c

For the graph of a differentiable function, the arc length is $L=\int \sqrt{1+(y^{\prime }{)}^2}dx$. Since $y^{\prime }=2x$,Use the hyperbolic substitution $2x=\sinh t$: then $dx=\frac{1}{2}\cosh tdt$ and $\sqrt{1+4x^2}=\cosh t$. The hyperbolic functions ${\cosh }^{2}t=(1+\cosh 2t)/2$ yieldsHere $\mathrm{arsinh}t=\ln (t+\sqrt{1+t^2})$ is the inverse hyperbolic sine.