Answer
$$ - \frac{{\sqrt {3 - 4{x^2}} }}{{3x}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{{x^2}\sqrt {3 - 4{x^2}} }}} dx,\,\,\,\,u = 2x \cr
& = \int {\frac{4}{{4{x^2}\sqrt {3 - {{\left( {2x} \right)}^2}} }}} dx \cr
& {\text{Using the given substitution}} \cr
& u = 2x,\,\,\,\,\,\,\,du = 2dx,\,\,\,\,\,\,dx = \frac{1}{2}du \cr
& {\text{write in terms of }}u \cr
& \int {\frac{4}{{{{\left( {2x} \right)}^2}\sqrt {3 - {{\left( {2x} \right)}^2}} }}} dx = \int {\frac{4}{{{{\left( {2x} \right)}^2}\sqrt {3 - {u^2}} }}} \left( {\frac{1}{2}du} \right) \cr
& = \int {\frac{2}{{{u^2}\sqrt {{{\left( {\sqrt 3 } \right)}^2} - {u^2}} }}} du \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{By formula 83}} \cr
& \left( {83} \right):\,\,\,\,\,\,\int {\frac{{du}}{{{u^2}\sqrt {{a^2} - {u^2}} }} = - \frac{{\sqrt {{a^2} - {u^2}} }}{{{a^2}u}}} + C \cr
& {\text{take }}a = \sqrt 3 \cr
& 2\int {\frac{1}{{{u^2}\sqrt {{{\left( {\sqrt 3 } \right)}^2} - {u^2}} }}} du = - \frac{{2\sqrt {{{\left( {\sqrt 3 } \right)}^2} - {u^2}} }}{{{{\left( {\sqrt 3 } \right)}^2}u}} + C \cr
& {\text{simplifying}} \cr
& = - \frac{{2\sqrt {3 - {u^2}} }}{{3u}} + C \cr
& {\text{write in terms of }}x{\text{, and replace }}2x{\text{ for }}u \cr
& = - \frac{{2\sqrt {3 - {{\left( {2x} \right)}^2}} }}{{3\left( {2x} \right)}} + C \cr
& = - \frac{{\sqrt {3 - 4{x^2}} }}{{3x}} + C \cr} $$
