Answer
$$\frac{2}{{\sqrt 6 }}{\tan ^{ - 1}}\sqrt {\frac{{2x - 3}}{3}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{x\sqrt {4x - 6} }}} \cr
& {\text{A matching integral in a table of integrals at the end of the book is the }} \cr
& {\text{formula }}88 \cr
& \int {\frac{{dx}}{{x\sqrt {ax - b} }} = \frac{2}{{\sqrt b }}{{\tan }^{ - 1}}\sqrt {\frac{{ax - b}}{b}} + C} \cr
& {\text{ }}a = 4,{\text{ }}b = 6 \cr
& then \cr
& \int {\frac{{dx}}{{x\sqrt {4x - 6} }}} = \frac{2}{{\sqrt 6 }}{\tan ^{ - 1}}\sqrt {\frac{{4x - 6}}{6}} + C \cr
& {\text{Simplify}} \cr
& \int {\frac{{dx}}{{x\sqrt {4x - 6} }}} = \frac{2}{{\sqrt 6 }}{\tan ^{ - 1}}\sqrt {\frac{{2x - 3}}{3}} + C \cr} $$
