Chapter 7: Transcendental Functions - Section 7.3 - Exponential Functions - Exercises 7.3 - Page 391: 107

\begin{aligned} \int_{1}^{\ln x} \frac{1}{t} d t =\ln (\ln x), \ \ \ \ \ \ \ \ x>1 \end{aligned}

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