Answer
$$\frac{{3\sqrt {{x^2} - 4} }}{x} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{ - 12}}{{{x^2}\sqrt {4 - {x^2}} }}} dx \cr
& {\text{Let }}x = 2\sin \theta ,{\text{ }}dx = 2\cos \theta d\theta \cr
& {\text{Substitute}} \cr
& \int {\frac{{ - 12}}{{{x^2}\sqrt {4 - {x^2}} }}} dx = \int {\frac{{ - 12}}{{{{\left( {2\sin \theta } \right)}^2}\sqrt {4 - {{\left( {2\sin \theta } \right)}^2}} }}\left( {2\cos \theta } \right)d\theta } \cr
& = \int {\frac{{ - 12}}{{4{{\sin }^2}\theta \sqrt {4 - 4{{\sin }^2}\theta } }}\left( {2\cos \theta } \right)d\theta } \cr
& = \int {\frac{{ - 12}}{{4{{\sin }^2}\theta \sqrt {4\left( {1 - {{\sin }^2}\theta } \right)} }}\left( {2\cos \theta } \right)d\theta } \cr
& = \int {\frac{{ - 12}}{{4{{\sin }^2}\theta \sqrt {4{{\cos }^2}\theta } }}\left( {2\cos \theta } \right)d\theta } \cr
& = \int {\frac{{ - 12}}{{4{{\sin }^2}\theta \left( {2\cos \theta } \right)}}\left( {2\cos \theta } \right)d\theta } \cr
& = - 3\int {\frac{1}{{{{\sin }^2}\theta }}d\theta } \cr
& = - 3\int {{{\csc }^2}\theta } d\theta \cr
& {\text{Integrate by basic rules}} \cr
& = - 3\left( { - \cot \theta } \right) + C \cr
& = 3\cot \theta + C \cr
& {\text{We know that }}x = 2\sin \theta ,{\text{ sin}}\theta = \frac{x}{2},{\text{ cot}}\theta = \frac{{\sqrt {{x^2} - 4} }}{x} \cr
& = 3\left( {\frac{{\sqrt {{x^2} - 4} }}{x}} \right) + C \cr
& = \frac{{3\sqrt {{x^2} - 4} }}{x} + C \cr} $$
