Answer
$$ - \frac{{\sqrt {9 - 4x} }}{x} - \frac{2}{3}\ln \left| {\frac{{\sqrt {9 - 4x} - 3}}{{\sqrt {9 - 4x} + 3}}} \right| + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sqrt {9 - 4x} }}{{{x^2}}}} dx \cr
& \int {\frac{{\sqrt { - 4x + 9} }}{{{x^2}}}} dx \cr
& {\text{use the table of integrals from the book}} \cr
& {\text{by the formula 30}}:\,\,\,\,\,\,\,\int {\frac{{\sqrt {ax + b} }}{{{x^2}}}dx = - \frac{{\sqrt {ax + b} }}{x} + \frac{a}{2}\int {\frac{{dx}}{{x\sqrt {ax + b} }}} } + C \cr
& {\text{setting }}a = - 4{\text{ and }}b = 9{\text{ then}} \cr
& \int {\frac{{\sqrt { - 4x + 9} }}{{{x^2}}}} dx = - \frac{{\sqrt {9 - 4x} }}{x} + \frac{{ - 4}}{2}\int {\frac{{dx}}{{x\sqrt {9 - 4x} }}} + C \cr
& {\text{simplifying}} \cr
& = - \frac{{\sqrt {9 - 4x} }}{x} - 2\int {\frac{{dx}}{{x\sqrt {9 - 4x} }}} + C \cr
& {\text{use }}\int {\frac{{dx}}{{x\sqrt {ax + b} }}} = \frac{1}{{\sqrt b }}\ln \left| {\frac{{\sqrt {ax + b} - \sqrt b }}{{\sqrt {ax + b} + \sqrt b }}} \right| + C \cr
& = - \frac{{\sqrt {9 - 4x} }}{x} - \frac{2}{{\sqrt 9 }}\ln \left| {\frac{{\sqrt {9 - 4x} - \sqrt 9 }}{{\sqrt {9 - 4x} + \sqrt 9 }}} \right| + C \cr
& = - \frac{{\sqrt {9 - 4x} }}{x} - \frac{2}{3}\ln \left| {\frac{{\sqrt {9 - 4x} - 3}}{{\sqrt {9 - 4x} + 3}}} \right| + C \cr} $$
