= Solution
For $x>0$, $\sqrt{1+x^2}$ increases from $1$, while $2/\sqrt{1+x^2}$ decreases from $2$. They meet at $(1,\sqrt2)$, after which their order reverses. In particular, their positive separation on $0<x<1$ is $(1-x^2)/\sqrt{1+x^2}$.
\Image[/past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2018-hyperbolic-area.png]
{title=The two curves and their enclosed positive area on $0<x<1$, with crossing at $(1,\sqrt2)$}
Use the <hyperbolic substitution> $x=\sinh t$, $dx=\cosh t\,dt$. With $T=\operatorname{arsinh}1$, the <definite integral> for the area becomes
$$
A=\int_0^T(2-\cosh^2t)\,dt
=\left[\frac32t-\frac14\sinh2t\right]_0^T.
$$
Since $\sinh T=1$, $\cosh T=\sqrt2$ and $T=\log(1+\sqrt2)$, the answer is
$$
\boxed{A=\frac{3\log(1+\sqrt2)-\sqrt2}{2}.}
$$
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