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

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.

A single formula valid on both real components isLet $s=\sqrt{x^2-1}>0$. Since $x+s\ne 0$,Equivalently, $R(x)=\mathrm{arcosh}x$ for $x>1$ and $R(x)=-\mathrm{arcosh}(-x)$ for $x<-1$. The graph is odd for this choice of constants, increases on both components, and approaches zero with vertical tangent as $x\to 1^{+}$ or $x\to -1^{-}$. Its far-field behaviors are $R(x)\sim \ln (2x)$ on the right and $R(x)\sim -\ln (-2x)$ on the left. The interval $[-1,1]$ is excluded. Independent additive constants could again be assigned to the two components. The right panel sketches these real inverse-hyperbolic antiderivative branches.