Answer
$$ - \frac{{\sqrt {6x - {x^2}} }}{{3x}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{x\sqrt {6x - {x^2}} }}} dx \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{Rewrite the integrand}} \cr
& \int {\frac{1}{{x\sqrt {2\left( 3 \right)x - {x^2}} }}} dx = \int {\frac{1}{{x\sqrt {2\left( 3 \right)x - {x^2}} }}} dx \cr
& {\text{The integrand has a expression in the form }}\sqrt {{a^2} - {u^2}} {} \cr
& {\text{Use formula 117}} \cr
& \left( {117} \right):\,\,\,\,\int {\frac{{du}}{{u\sqrt {2au - {u^2}} }}du} = - \frac{{\sqrt {2au - {u^2}} }}{{au}} + C \cr
& {\text{let }}u = x,\,\,\,a = 3 \cr
& \int {\frac{1}{{x\sqrt {2\left( 3 \right)x - {x^2}} }}} dx = - \frac{{\sqrt {2\left( 3 \right)x - {x^2}} }}{{3x}} + C \cr
& {\text{simplifying}} \cr
& \int {\frac{1}{{x\sqrt {6x - {x^2}} }}} dx = - \frac{{\sqrt {6x - {x^2}} }}{{3x}} + C \cr} $$
