Substitute \(z=\sqrt{x^2-1}\), so that \(x^2=z^2+1\). Then \(\mathrm dz=\dfrac x{\sqrt{x^2-1}}\,\mathrm dx\). You have
\(\displaystyle\int\frac{\mathrm dx}{x\sqrt{x^2-1}}=\int\frac x{x^2\sqrt{x^2-1}}\,\mathrm dx=\int\frac{\mathrm dz}{z^2+1}\)
This is a standard integral, and the antiderivative is
\(\arctan z+C\)
(you can verify with a trig sub of \(z=\tan u\), for instance). Transforming back to a function of \(x\), you get
\(\displaystyle\int\frac{\mathrm dx}{x\sqrt{x^2-1}}=\arctan(\sqrt{x^2-1})+C\)
