Answer
$$\frac{1}{{\sqrt 3 }}\ln \left| {\frac{{\sqrt {4x + 3} - \sqrt 3 }}{{\sqrt {4x + 3} + \sqrt 3 }}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{x\sqrt {4x + 3} }}} \cr
& {\text{write in terms of }}u.{\text{ let }}u = x,\,\,\,du = dx \cr
& = \int {\frac{{du}}{{u\sqrt {3 + 4u} }}} \cr
& \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral by the formula }} \cr
& \int {\frac{{du}}{{u\sqrt {a + bu} }}} = \frac{1}{{\sqrt a }}\ln \left| {\frac{{\sqrt {a + bu} - \sqrt a }}{{\sqrt {a + bu} + \sqrt a }}} \right| + C,\,\,\,{\text{with }}a > 0 \cr
& {\text{We have }}a = 3,\,\,\,b = 4.{\text{ Then}} \cr
& \int {\frac{{du}}{{u\sqrt {3 + 4u} }}} = \frac{1}{{\sqrt 3 }}\ln \left| {\frac{{\sqrt {3 + 4u} - \sqrt 3 }}{{\sqrt {3 + 4u} + \sqrt 3 }}} \right| + C \cr
& \cr
& {\text{Write in terms of }}x{\text{ replace }}x{\text{ for }}u \cr
& \int {\frac{{dx}}{{x\sqrt {4x + 3} }}} = \frac{1}{{\sqrt 3 }}\ln \left| {\frac{{\sqrt {4x + 3} - \sqrt 3 }}{{\sqrt {4x + 3} + \sqrt 3 }}} \right| + C \cr} $$
