Chapter 4 - Integrals - 4.3 The Fundamental Theorem of Calculus - 4.3 Exercises - Page 330: 79

$$ \int_{1}^{9} \frac{1}{2 x} d x = \ln 3 $$

$$ \begin{aligned} \int_{1}^{9} \frac{1}{2 x} d x & =\frac{1}{2} \int_{1}^{9} \frac{1}{x} d x\\ &=\frac{1}{2}[\ln |x|]_{1}^{9} \\ &=\frac{1}{2}(\ln 9-\ln 1)\\ &=\frac{1}{2} \ln 9-0\\ &=\ln 9^{1 / 2}\\ &=\ln 3 \end{aligned} $$