Answer
$$\frac{1}{{\sqrt 2 }}se{c^{ - 1}}\left| {\frac{{5x}}{{\sqrt 2 }}} \right| + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{x\sqrt {25{x^2} - 2} }}} \cr
& {\text{write }}25{x^2} - 2{\text{ as }}{\left( {5x} \right)^2} - {\left( {\sqrt 2 } \right)^2} \cr
& {\text{use the substitution method}}{\text{.}} \cr
& u = 5x,{\text{ so that }}du = 5dx,\,\,\,\,\,\,dx = \frac{{du}}{5} \cr
& {\text{then}} \cr
& \int {\frac{{dx}}{{x\sqrt {25{x^2} - 2} }}} = \int {\frac{{du/5}}{{x\sqrt {{u^2} - {{\left( {\sqrt 2 } \right)}^2}} }}} \cr
& = \int {\frac{{du}}{{5x\sqrt {{u^2} - {{\left( {\sqrt 2 } \right)}^2}} }}} \cr
& = \int {\frac{{du}}{{u\sqrt {{u^2} - {{\left( {\sqrt 2 } \right)}^2}} }}} \cr
& {\text{intgrate by using the formula }}\int {\frac{{du}}{{u\sqrt {{u^2} - {a^2}} }} = \frac{1}{a}se{c^{ - 1}}\left| {\frac{u}{a}} \right| + C\,\,\,\left( {{\text{see page 419}}} \right)} \cr
& {\text{with }}a = \sqrt 2 \cr
& = \int {\frac{{du}}{{u\sqrt {{u^2} - {{\left( {\sqrt 2 } \right)}^2}} }}} = \frac{1}{{\sqrt 2 }}se{c^{ - 1}}\left| {\frac{u}{{\sqrt 2 }}} \right| + C \cr
& {\text{write in terms of }}x;{\text{ replace }}5x{\text{ for }}u \cr
& = \frac{1}{{\sqrt 2 }}se{c^{ - 1}}\left| {\frac{{5x}}{{\sqrt 2 }}} \right| + C \cr} $$
