Answer
$$\frac{x}{2}\sqrt {{x^2} - 3} - \frac{3}{2}\ln \left| {x + \sqrt {{x^2} - 3} } \right| + C\,\,\,$$
Work Step by Step
$$\eqalign{
& \int {\sqrt {{x^2} - 3} } dx \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{Rewrite the integrand}} \cr
& \int {\sqrt {{x^2} - 3} } dx = \int {\sqrt {{x^2} - {{\left( {\sqrt 3 } \right)}^2}} } dx \cr
& {\text{The integrand has a expression in the form }}\sqrt {{u^2} - {a^2}} {\text{}}{\text{}} \cr
& {\text{Use formula 73}} \cr
& \left( {73} \right):\,\,\,\,\int {\sqrt {{u^2} - {a^2}} du} = \frac{u}{2}\sqrt {{u^2} - {a^2}} - \frac{{{a^2}}}{2}\ln \left| {u + \sqrt {{u^2} - {a^2}} } \right| + C\,\,\, \cr
& {\text{let }}u = x,\,\,\,a = \sqrt 3 \cr
& \int {\sqrt {{x^2} - 3} } dx = \frac{x}{2}\sqrt {{x^2} - {{\left( {\sqrt 3 } \right)}^2}} - \frac{{{{\left( {\sqrt 3 } \right)}^2}}}{2}\ln \left| {x + \sqrt {{x^2} - {{\left( {\sqrt 3 } \right)}^2}} } \right| + C\,\,\, \cr
& {\text{simplifying}} \cr
& \int {\sqrt {{x^2} - 3} } dx = \frac{x}{2}\sqrt {{x^2} - 3} - \frac{3}{2}\ln \left| {x + \sqrt {{x^2} - 3} } \right| + C\,\,\, \cr} $$
