Answer
$$\frac{1}{2}{\sec ^{ - 1}}\left| {\frac{{\sqrt 5 x}}{2}} \right| + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{x\sqrt {5{x^2} - 4} }}} \cr
& {\text{write }}5{x^2} - 4{\text{ as }}{\left( {\sqrt 5 x} \right)^2} - {\left( 2 \right)^2} \cr
& = \int {\frac{{dx}}{{x\sqrt {{{\left( {\sqrt 5 x} \right)}^2} - {{\left( 2 \right)}^2}} }}} \cr
& {\text{use the substitution method}}{\text{.}} \cr
& u = \sqrt 5 x,{\text{ so that }}du = \sqrt 5 dx,\,\,\,\,\,\,dx = \frac{{du}}{{\sqrt 5 }} \cr
& {\text{then}} \cr
& \int {\frac{{dx}}{{x\sqrt {{{\left( {\sqrt 5 x} \right)}^2} - {{\left( 2 \right)}^2}} }}} = \int {\frac{{du/\sqrt 5 }}{{x\sqrt {{u^2} - {2^2}} }}} \cr
& = \int {\frac{{du}}{{\sqrt 5 x\sqrt {{u^2} - {2^2}} }}} \cr
& = \int {\frac{{du}}{{u\sqrt {{u^2} - {2^2}} }}} \cr
& {\text{intgrate by using the formula }}\int {\frac{{du}}{{u\sqrt {{u^2} - {a^2}} }} = \frac{1}{a}se{c^{ - 1}}\left| {\frac{u}{a}} \right| + C\,\,\,\left( {{\text{see page 419}}} \right)} \cr
& {\text{with }}a = 2 \cr
& = \int {\frac{{du}}{{u\sqrt {{u^2} - {2^2}} }}} = \frac{1}{2}se{c^{ - 1}}\left| {\frac{u}{2}} \right| + C \cr
& {\text{write in terms of }}x;{\text{ replace }}\sqrt 5 x{\text{ for }}u \cr
& = \frac{1}{2}{\sec ^{ - 1}}\left| {\frac{{\sqrt 5 x}}{2}} \right| + C \cr} $$
