Answer
$$ = 3\ln 2$$
Work Step by Step
$$\eqalign{
& \int_1^2 {\frac{3}{t}} dt \cr
& 3\int_1^2 {\frac{1}{t}} dt \cr
& {\text{recall that }}\int {\frac{1}{x}} dx = \ln \left| x \right| + C,{\text{ so for }}t \cr
& \int_1^2 {\frac{3}{t}} dt = 3\left. {\left( {\ln \left| t \right|} \right)} \right|_1^2 \cr
& {\text{using The Fundamental Theorem}} \cr
& = 3\left( {\ln \left| 2 \right|} \right) - 3\left( {\ln \left| 1 \right|} \right) \cr
& {\text{simplify}} \cr
& = 3\ln \left( 2 \right) - 3\left( 0 \right) \cr
& = 3\ln 2 \cr} $$
