Answer
$\ln \left| {\frac{{3x}}{{\sqrt {9{x^2} + 1} + 1}}} \right| + C$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{x\sqrt {9{x^2} + 1} }}} dx \cr
& {\text{Rewrite}} \cr
& \int {\frac{1}{{x\sqrt {9{x^2} + 1} }}} dx = \int {\frac{1}{{x\sqrt {{{\left( {3x} \right)}^2} + 1} }}} dx \cr
& {\text{Let }}u = 3x,{\text{ and }}du = 3dx,{\text{ }}dx = \frac{1}{3}du \cr
& \int {\frac{1}{{x\sqrt {{{\left( {3x} \right)}^2} + 1} }}} dx = \int {\frac{1}{{\left( {u/3} \right)\sqrt {{u^2} + 1} }}\left( {\frac{1}{3}} \right)} du \cr
& = \int {\frac{1}{{u\sqrt {{u^2} + 1} }}} du \cr
& {\text{Using trigonometric substitution}} \cr
& {\text{ }}u = \tan \theta ,{\text{ }}du = {\sec ^2}\theta d\theta \cr
& \int {\frac{1}{{u\sqrt {{u^2} + 1} }}} du = \int {\frac{1}{{\tan \theta \sqrt {{{\tan }^2}\theta + 1} }}} \left( {{{\sec }^2}\theta } \right)d\theta \cr
& = \int {\frac{1}{{\tan \theta \sqrt {{{\sec }^2}\theta } }}} \left( {{{\sec }^2}\theta } \right)d\theta \cr
& = \int {\frac{{\sec \theta }}{{\tan \theta }}} d\theta \cr
& = \int {\left( {\frac{1}{{\cos \theta }}} \right)\left( {\frac{{\cos \theta }}{{\sin \theta }}} \right)} d\theta \cr
& = \int {\frac{1}{{\sin \theta }}} d\theta \cr
& = \int {\csc \theta } d\theta \cr
& {\text{Integrating}} \cr
& = - \ln \left| {\csc \theta + \cot \theta } \right| + C \cr
& {\text{Write in terms of }}u \cr
& \cot \theta = \frac{1}{u}{\text{ and }}\csc \theta = \frac{{\sqrt {{u^2} + 1} }}{u} \cr
& = - \ln \left| {\frac{{\sqrt {{u^2} + 1} }}{u} + \frac{1}{u}} \right| + C \cr
& = - \ln \left| {\frac{{\sqrt {{u^2} + 1} + 1}}{u}} \right| + C \cr
& or \cr
& = \ln \left| {\frac{u}{{\sqrt {{u^2} + 1} + 1}}} \right| + C \cr
& {\text{Write in terms of }}x,{\text{ substitute back }}u = 3x \cr
& = \ln \left| {\frac{{3x}}{{\sqrt {{{\left( {3x} \right)}^2} + 1} + 1}}} \right| + C \cr
& = \ln \left| {\frac{{3x}}{{\sqrt {9{x^2} + 1} + 1}}} \right| + C \cr} $$
