Answer
$$2\ln \left| {\frac{{\sqrt {4x + 1} + 1}}{{\sqrt {4x + 1} - 1}}} \right| - \frac{{4\sqrt {4x + 1} }}{{{{\left( {1 - \sqrt {4x + 1} } \right)}^2}}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{{x^2}\sqrt {4x + 1} }}} dx \cr
& {\text{Let }}{u^2} = 4x + 1,{\text{ }}{x^2} = \frac{{{{\left( {{u^2} - 1} \right)}^2}}}{{16}},{\text{ }}2udu = 4dx,{\text{ }}dx = \frac{1}{2}udu \cr
& {\text{Substituting}} \cr
& \int {\frac{1}{{{x^2}\sqrt {4x + 1} }}} dx = \int {\frac{{16}}{{{{\left( {{u^2} - 1} \right)}^2}\sqrt {{u^2}} }}\left( {\frac{1}{2}u} \right)} du \cr
& = \int {\frac{8}{{u{{\left( {{u^2} - 1} \right)}^2}}}\left( u \right)} du \cr
& = 8\int {\frac{1}{{{{\left( {{u^2} - 1} \right)}^2}}}} du \cr
& {\text{Difference of two squares}} \cr
& = 8\int {\frac{1}{{{{\left[ {\left( {u + 1} \right)\left( {u - 1} \right)} \right]}^2}}}} du \cr
& = 8\int {\frac{1}{{{{\left( {u + 1} \right)}^2}{{\left( {u - 1} \right)}^2}}}} du \cr
& {\text{Decomposing into partial fractions we obtain}} \cr
& = 8\int {\left[ {\frac{1}{{4\left( {u + 1} \right)}} + \frac{1}{{4{{\left( {u + 1} \right)}^2}}} - \frac{1}{{4\left( {u - 1} \right)}} + \frac{1}{{4{{\left( {u - 1} \right)}^2}}}} \right]} du \cr
& {\text{Integrating}} \cr
& = 8\left( {\frac{1}{4}\ln \left| {u + 1} \right| - \frac{1}{{4\left( {u + 1} \right)}} - \frac{1}{4}\ln \left| {u - 1} \right| - \frac{1}{{4\left( {u - 1} \right)}}} \right) + C \cr
& = 8\left( {\frac{1}{4}\ln \left| {\frac{{u + 1}}{{u - 1}}} \right| - \frac{u}{{2{{\left( {1 - u} \right)}^2}}}} \right) + C \cr
& = 2\ln \left| {\frac{{u + 1}}{{u - 1}}} \right| - \frac{{4u}}{{{{\left( {1 - u} \right)}^2}}} + C \cr
& {\text{Write in terms of }}x,{\text{ }}{u^2} = 4x + 1 \to u = \sqrt {4x + 1} \cr
& = 2\ln \left| {\frac{{\sqrt {4x + 1} + 1}}{{\sqrt {4x + 1} - 1}}} \right| - \frac{{4\sqrt {4x + 1} }}{{{{\left( {1 - \sqrt {4x + 1} } \right)}^2}}} + C \cr} $$
