Answer
$$12\ln \left| x \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{12}}{x}} dx \cr
& {\text{pull out the constant 12}} \cr
& = 12\int {\frac{1}{x}} dx \cr
& {\text{use the basic formula of the integration }}\int {\frac{1}{x}} dx = \ln \left| x \right| + C \cr
& 12\int {\frac{1}{x}} dx = 12\left( {\ln \left| x \right|} \right) + C \cr
& {\text{simplify}} \cr
& = 12\ln \left| x \right| + C \cr} $$
