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