Answer
$$ - \frac{{2\sqrt {x - 5{x^2}} }}{x} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{x\sqrt {x - 5{x^2}} }}dx} \cr
& {\text{write the integrand as}} \cr
& = \int {\frac{{\sqrt 5 }}{{\sqrt 5 x\sqrt {x - {{\left( {\sqrt 5 x} \right)}^2}} }}dx} \cr
& {\text{Make an appropiate }}u{\text{ - substitution }} \cr
& u = \sqrt 5 x,\,\,\,\,\,\,\,du = \sqrt 5 dx,\,\,\,\,\,\,dx = \frac{{du}}{{\sqrt 5 }} \cr
& {\text{write in terms of }}u \cr
& \int {\frac{{\sqrt 5 }}{{\sqrt 5 x\sqrt {x - {{\left( {\sqrt 5 x} \right)}^2}} }}dx} = \int {\frac{{\sqrt 5 }}{{u\sqrt {\frac{u}{{\sqrt 5 }} - {u^2}} }}\left( {\frac{{du}}{{\sqrt 5 }}} \right)} \cr
& = \frac{1}{{\sqrt 5 }}\int {\frac{1}{{u\sqrt {\frac{1}{{\sqrt 5 }}u - {u^2}} }}du} \cr
& = \frac{1}{{\sqrt 5 }}\int {\frac{1}{{u\sqrt {2\left( {\frac{1}{{2\sqrt 5 }}} \right)u - {u^2}} }}du} \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{By formula 117}} \cr
& \left( {117} \right):\,\,\,\,\,\,\int {\frac{{du}}{{u\sqrt {2au - {u^2}} }} = } - \frac{{\sqrt {2au - {u^2}} }}{{au}} + C \cr
& {\text{take }}a = \frac{1}{{2\sqrt 5 }},\,\, \cr
& \frac{1}{{\sqrt 5 }}\int {\frac{1}{{u\sqrt {2\left( {\frac{1}{{2\sqrt 5 }}} \right)u - {u^2}} }}du} = - \frac{{\sqrt {2\left( {1/2\sqrt 5 } \right)u - {u^2}} }}{{\left( {1/2\sqrt 5 } \right)u}} + C \cr
& {\text{simplifying}} \cr
& = - \frac{{2\sqrt {\frac{1}{{\sqrt 5 }}u - {u^2}} }}{{\left( {1/2\sqrt 5 } \right)u}} + C \cr
& {\text{write in terms of }}x{\text{; replace }}\sqrt 5 x{\text{ for }}u \cr
& = - \frac{{2\sqrt {\frac{1}{{\sqrt 5 }}\left( {\sqrt 5 x} \right) - {{\left( {\sqrt 5 x} \right)}^2}} }}{{\left( {1/2\sqrt 5 } \right)\left( {\sqrt 5 x} \right)}} + C \cr
& = - \frac{{2\sqrt {x - 5{x^2}} }}{x} + C \cr} $$
