Answer
$$ - \frac{{\sqrt {4 - 9{x^2}} }}{x} - 3{\sin ^{ - 1}}\frac{{3x}}{2} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sqrt {4 - 9{x^2}} }}{{{x^2}}}dx} \cr
& {\text{write the integrand as}} \cr
& = 9\int {\frac{{\sqrt {{{\left( 2 \right)}^2} - {{\left( {3x} \right)}^2}} }}{{{{\left( {3x} \right)}^2}}}dx} \cr
& {\text{Make an appropiate }}u{\text{ - substitution }} \cr
& u = 3x,\,\,\,\,\,\,\,du = 3dx,\,\,\,\,\,\,dx = \frac{{du}}{3} \cr
& {\text{write in terms of }}u \cr
& = 9\int {\frac{{\sqrt {{{\left( 2 \right)}^2} - {{\left( {3x} \right)}^2}} }}{{{{\left( {3x} \right)}^2}}}dx} = 9\int {\frac{{\sqrt {{{\left( 2 \right)}^2} - {u^2}} }}{{{u^2}}}\left( {\frac{{du}}{3}} \right)} \cr
& = 3\int {\frac{{\sqrt {{{\left( 2 \right)}^2} - {u^2}} }}{{{u^2}}}du} \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{By formula 80}} \cr
& \left( {80} \right):\,\,\,\,\,\,\int {\frac{{\sqrt {{a^2} - {u^2}} du}}{{{u^2}}} = } - \frac{{\sqrt {{a^2} - {u^2}} }}{u} - {\sin ^{ - 1}}\frac{u}{a} + C \cr
& {\text{take }}a = 2,\,\, \cr
& 3\int {\frac{{\sqrt {{{\left( 2 \right)}^2} - {u^2}} }}{{{u^2}}}du} = - \frac{{3\sqrt {{{\left( 2 \right)}^2} - {u^2}} }}{u} - 3{\sin ^{ - 1}}\frac{u}{2} + C \cr
& {\text{simplifying}} \cr
& = - \frac{{3\sqrt {4 - {u^2}} }}{u} - 3{\sin ^{ - 1}}\frac{u}{2} + C \cr
& {\text{write in terms of }}x{\text{; replace }}3x{\text{ for }}u \cr
& = - \frac{{3\sqrt {4 - {{\left( {3x} \right)}^2}} }}{{3x}} - 3{\sin ^{ - 1}}\frac{{3x}}{2} + C \cr
& = - \frac{{\sqrt {4 - 9{x^2}} }}{x} - 3{\sin ^{ - 1}}\frac{{3x}}{2} + C \cr} $$
