# Chapter 5 - Integration - 5.3 Fundamental Theorem of Calculus - 5.3 Exercises - Page 374: 45

$$= 3\ln 2$$

#### 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}

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