Past exam of the mathematical methods II course of the University of Oxford / 2018 / 4 / Solution

For $x>0$, $\sqrt{1+x^2}$ increases from $1$, while $2/\sqrt{1+x^2}$ decreases from $2$. They meet at $(1,\sqrt{2})$, after which their order reverses. In particular, their positive separation on $0<x<1$ is $(1-x^2)/\sqrt{1+x^2}$.

Use the hyperbolic substitution $x=\sinh t$, $dx=\cosh tdt$. With $T=\mathrm{arsinh}1$, the definite integral for the area becomesSince $\sinh T=1$, $\cosh T=\sqrt{2}$ and $T=\log (1+\sqrt{2})$, the answer is