Answer
$$\frac{{\sqrt {{x^2} - 2} }}{{2x}} + C\,\,\,\,\,\,$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{{x^2}\sqrt {{x^2} - 2} }}} dx \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{Rewrite the integrand}} \cr
& \int {\frac{1}{{{x^2}\sqrt {{x^2} - 2} }}} dx = \int {\frac{1}{{{x^2}\sqrt {{x^2} - {{\left( {\sqrt 2 } \right)}^2}} }}} dx \cr
& {\text{The integrand has a expression in the form }}\sqrt {{u^2} - {a^2}} {} \cr
& {\text{Use formula 90}} \cr
& \left( {90} \right):\,\,\,\,\int {\frac{{{u^2}}}{{\sqrt {{u^2} \pm {a^2}} }}du} = \mp \frac{{\sqrt {{u^2} + {a^2}} }}{{{a^2}u}} + C\,\,\, \cr
& {\text{let }}u = x,\,\,\,a = \sqrt 2 \cr
& \int {\frac{1}{{{x^2}\sqrt {{x^2} - {{\left( {\sqrt 2 } \right)}^2}} }}} dx = \frac{{\sqrt {{x^2} - {{\left( {\sqrt 2 } \right)}^2}} }}{{{{\left( {\sqrt 2 } \right)}^2}x}} + C\,\,\,\,\,\, \cr
& {\text{simplifying}} \cr
& \int {\frac{1}{{{x^2}\sqrt {{x^2} - 2} }}} dx = \frac{{\sqrt {{x^2} - 2} }}{{2x}} + C \cr} $$
