Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.6 Logarithmic And Other Functions Defined By Integrals - Exercises Set 6.6 - Page 460: 19

$\bf{True}$

For $a \gt0$, we have: $\int_{1}^{1/a} \dfrac{1}{t} \ dt=[\ln t]_{1}^{1/a}=[-\ln (a)-0]=-\ln (a)$ Also, $- \int_{1}^{a} \dfrac{1}{t} \ dt=-[\ln t]_{1}^{a}=- [\ln (a)-0]=-\ln (a)$ Hence, the given statement is $\bf{True}$.