Answer
$$ - \frac{1}{{12}}\ln \left| {\frac{{12 + \sqrt {144 - {x^2}} }}{x}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{x\sqrt {144 - {x^2}} }}} \cr
& u = x,{\text{ }}du = dx,{\text{ }}a = 12 \cr
& \int {\frac{{dx}}{{x\sqrt {144 - {x^2}} }}} = \int {\frac{{du}}{{u\sqrt {{a^2} - {u^2}} }}} \cr
& {\text{a matching integral in a table of integrals at the end of the book is the }} \cr
& {\text{formula 62}} \cr
& \int {\frac{{du}}{{u\sqrt {{a^2} - {u^2}} }}} = - \frac{1}{a}\ln \left| {\frac{{a + \sqrt {{a^2} - {u^2}} }}{u}} \right| + C \cr
& u = x,{\text{ }}a = 12 \cr
& = - \frac{1}{{12}}\ln \left| {\frac{{12 + \sqrt {144 - {x^2}} }}{x}} \right| + C \cr} $$
