Answer
$${\tan ^{ - 1}}\frac{{\sqrt {3x - 4} }}{2} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{x\sqrt {3x - 4} }}} dx \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{Rewrite the integrand}} \cr
& = \int {\frac{1}{{x\sqrt { - 4 + 3x} }}} dx \cr
& {\text{The integrand has a expression in the form }}\sqrt {a + bu} {\text{}}{\text{}} \cr
& {\text{Use formula 108}} \cr
& \left( {108} \right):\,\,\,\,\int {\frac{{du}}{{u\sqrt {a + bu} }}} = \frac{1}{{\sqrt a }}\ln \left| {\frac{{\sqrt {a + bu} - \sqrt a }}{{\sqrt {a + bu} + \sqrt a }}} \right| + C\,\,\,\left( {a > 0} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{2}{{\sqrt { - a} }}{\tan ^{ - 1}}\sqrt {\frac{{a + bu}}{{ - a}}} + C\,\,\,\left( {a < 0} \right) \cr
& \int {\frac{1}{{x\sqrt { - 4 + 3x} }}} dx \cr
& {\text{let }}u = x,\,\,\,a = - 4{\text{ and }}b = 3 \cr
& a > 0,\,\, \cr
& = \frac{2}{{\sqrt { - a} }}{\tan ^{ - 1}}\sqrt {\frac{{a + bu}}{{ - a}}} + C\,\,\,\left( {a < 0} \right) \cr
& {\text{substituting }}a,{\text{ }}b{\text{ and }}x \cr
& = \frac{2}{{\sqrt { - \left( { - 4} \right)} }}{\tan ^{ - 1}}\sqrt {\frac{{ - 4 + 3x}}{{ - \left( { - 4} \right)}}} + C\,\,\,\left( {a < 0} \right) \cr
& {\text{simplifying}} \cr
& = \frac{2}{{\sqrt 4 }}{\tan ^{ - 1}}\sqrt {\frac{{ - 4 + 3x}}{4}} + C \cr
& = \frac{2}{2}{\tan ^{ - 1}}\sqrt {\frac{{3x - 4}}{4}} + C \cr
& = {\tan ^{ - 1}}\frac{{\sqrt {3x - 4} }}{2} + C \cr} $$
