Chapter 7: Transcendental Functions - Section 7.3 - Exponential Functions - Exercises 7.3 - Page 391: 109

\begin{aligned} \int_{1}^{1 / x} \frac{1}{t} d t =-\ln x, \ \ \ \ \ \ \ \ \ \ \ \ \ x>0 \end{aligned}

Given $$\int_{1}^{1 / x} \frac{1}{t} d t $$ So, we have \begin{aligned} I&=\int_{1}^{1 / x} \frac{1}{t} d t\\ &=[\ln |t|]_{1}^{1 / x}\\ &=\ln \left|\frac{1}{x}\right|-\ln 1\\ &=(\ln 1-\ln |x|)-\ln 1\\ &=-\ln x, \ \ \ \ \ \ \ \ \ \ \ \ \ x>0 \end{aligned}