Answer
$$ - \frac{1}{3}\ln \left| {\frac{{3 + \sqrt {4{x^2} + 9} }}{{2x}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{x\sqrt {4{x^2} + 9} }}} dx \cr
& {\text{Rewrite}} \cr
& \int {\frac{1}{{x\sqrt {4{x^2} + 9} }}} dx = \int {\frac{1}{{x\sqrt {{{\left( {2x} \right)}^2} + {{\left( 3 \right)}^2}} }}} dx \cr
& {\text{Let }}u = 2x,{\text{ and }}du = 2dx,{\text{ }}dx = \frac{1}{2}du \cr
& \int {\frac{1}{{x\sqrt {{{\left( {2x} \right)}^2} + {{\left( 3 \right)}^2}} }}} dx = \int {\frac{1}{{u/2\sqrt {{u^2} + {{\left( 3 \right)}^2}} }}\left( {\frac{1}{2}} \right)} du \cr
& = \int {\frac{1}{{u\sqrt {{u^2} + {3^2}} }}} du \cr
& {\text{Refer to the triangle below}} \cr
& \tan \theta = \frac{u}{3},{\text{ }}u = 3\tan \theta ,{\text{ }}du = 3{\sec ^2}\theta d\theta \cr
& \int {\frac{1}{{u\sqrt {{u^2} + {3^2}} }}} du = \int {\frac{1}{{3\tan \theta \sqrt {{{\left( {3\tan \theta } \right)}^2} + {3^2}} }}} \left( {3{{\sec }^2}\theta } \right)d\theta \cr
& = \int {\frac{{{{\sec }^2}\theta }}{{\tan \theta \sqrt {{3^2}\left( {{{\tan }^2}\theta + 1} \right)} }}} d\theta \cr
& = \int {\frac{{{{\sec }^2}\theta }}{{3\tan \theta \sqrt {{{\sec }^2}\theta } }}} d\theta \cr
& = \frac{1}{3}\int {\frac{{\sec \theta }}{{\tan \theta }}} d\theta \cr
& = \frac{1}{3}\int {\left( {\frac{1}{{\cos \theta }}} \right)\left( {\frac{{\cos \theta }}{{\sin \theta }}} \right)} d\theta \cr
& = \frac{1}{3}\int {\frac{1}{{\sin \theta }}} d\theta \cr
& = \frac{1}{3}\int {\csc \theta } d\theta \cr
& {\text{Integrating}} \cr
& = - \frac{1}{3}\ln \left| {\cot \theta + \csc \theta } \right| + C \cr
& {\text{Refer to the triangle}} \cr
& \cot \theta = \frac{3}{u}{\text{ and }}\csc \theta = \frac{{\sqrt {{u^2} + {3^2}} }}{u} \cr
& = - \frac{1}{3}\ln \left| {\frac{3}{u} + \frac{{\sqrt {{u^2} + {3^2}} }}{u}} \right| + C \cr
& = - \frac{1}{3}\ln \left| {\frac{{3 + \sqrt {{u^2} + {3^2}} }}{u}} \right| + C \cr
& {\text{Substitute back }}u = 2x \cr
& = - \frac{1}{3}\ln \left| {\frac{{3 + \sqrt {{{\left( {2x} \right)}^2} + {{\left( 3 \right)}^2}} }}{{2x}}} \right| + C \cr
& = - \frac{1}{3}\ln \left| {\frac{{3 + \sqrt {4{x^2} + 9} }}{{2x}}} \right| + C \cr} $$
