Answer
$$\ln \left| {\frac{{\sqrt {4x + 1} - 1}}{{\sqrt {4x + 1} + 1}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{x\sqrt {4x + 1} }}} dx \cr
& {\text{Let }}{u^2} = 4x + 1,{\text{ }}x = \frac{{{u^2} - 1}}{4},{\text{ }}2udu = 4dx,{\text{ }}dx = \frac{1}{2}udu \cr
& {\text{Substituting}} \cr
& = \int {\frac{4}{{\left( {{u^2} - 1} \right)\sqrt {{u^2}} }}\left( {\frac{1}{2}u} \right)} du \cr
& = \int {\frac{2}{{\left( {{u^2} - 1} \right)u}}\left( u \right)} du \cr
& = 2\int {\frac{1}{{{u^2} - 1}}} du \cr
& {\text{Integrate using the formula }}\int {\frac{{dx}}{{{x^2} - {a^2}}} = \frac{1}{{2a}}\ln \left| {\frac{{x - a}}{{x + a}}} \right| + C} \cr
& \left( {{\text{Page 503}}} \right): \cr
& 2\int {\frac{1}{{{u^2} - 1}}} du = 2\left( {\frac{1}{{2\left( 1 \right)}}\ln \left| {\frac{{u - 1}}{{u + 1}}} \right|} \right) + C \cr
& = \ln \left| {\frac{{u - 1}}{{u + 1}}} \right| + C \cr
& {\text{Write in terms of }}x,{\text{ }}{u^2} = 4x + 1 \to u = \sqrt {4x + 1} \cr
& = \ln \left| {\frac{{\sqrt {4x + 1} - 1}}{{\sqrt {4x + 1} + 1}}} \right| + C \cr} $$
