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