= Solution
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 $z=u+\sqrt{u^2-1}$,
$$
\frac{dz}{du}=1+\frac{u}{\sqrt{u^2-1}}=\frac z{\sqrt{u^2-1}},\qquad\frac{du}{\sqrt{u^2-1}}=\frac{dz}{z}.
$$
The lower endpoint corresponds to $z=1$, giving
$$
\boxed{F(x)=\ln\left(x+\sqrt{x^2-1}\right)=\operatorname{arcosh}x,\qquad x>1.}
$$
The limiting value at $x\downarrow1$ is zero. Here $\operatorname{arcosh}$ is the <inverse hyperbolic cosine>.
Back to article page
