Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 1 / 17 / c / i

At $u=1$, the integrand is unbounded but integrable: $1/\sqrt{u^2-1}\sim [2(u-1){]}^{-1/2}$. Thus the lower-limit integral is understood as an improper integral. With $z=u+\sqrt{u^2-1}$,The lower endpoint corresponds to $z=1$, givingThe limiting value at $x\downarrow 1$ is zero. Here $\mathrm{arcosh}$ is the inverse hyperbolic cosine.