Answer
$$\frac{1}{2}\ln \left| {2x + \sqrt {4{x^2} - 9} } \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{\sqrt {4{x^2} - 9} }}} dx,\,\,\,\,u = 2x \cr
& = \int {\frac{1}{{\sqrt {{{\left( {2x} \right)}^2} - {3^2}} }}} dx \cr
& {\text{Using the given substitution}} \cr
& u = 2x,\,\,\,\,\,\,\,du = 2dx,\,\,\,\,\,\,dx = \frac{1}{2}du \cr
& {\text{write in terms of }}u \cr
& \int {\frac{1}{{\sqrt {{{\left( {2x} \right)}^2} - {3^2}} }}} dx = \int {\frac{1}{{\sqrt {{u^2} - {3^2}} }}} \left( {\frac{1}{2}du} \right) \cr
& = \frac{1}{2}\int {\frac{1}{{\sqrt {{u^2} - {3^2}} }}} du \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{By formula 76}} \cr
& \left( {76} \right):\,\,\,\,\,\,\int {\frac{{du}}{{\sqrt {{u^2} - {a^2}} }} = \ln \left| {u + \sqrt {{u^2} - {a^2}} } \right| + C} \cr
& {\text{take }}a = 3 \cr
& \frac{1}{2}\int {\frac{1}{{\sqrt {{u^2} - {3^2}} }}} du = \frac{1}{2}\ln \left| {u + \sqrt {{u^2} - {3^2}} } \right| + C \cr
& {\text{simplifying}} \cr
& = \frac{1}{2}\ln \left| {u + \sqrt {{u^2} - 9} } \right| + C \cr
& {\text{write in terms of }}x{\text{, and replace }}2x{\text{ for }}u \cr
& = \frac{1}{2}\ln \left| {2x + \sqrt {{{\left( {2x} \right)}^2} - 9} } \right| + C \cr
& = \frac{1}{2}\ln \left| {2x + \sqrt {4{x^2} - 9} } \right| + C \cr} $$
