Easy enough

$\begin{aligned} I & = \int_a^b \frac {e^\frac xa - e^\frac bx}x dx \\ & = \blue{\int_a^b \frac {e^\frac xa}x dx} - \red{\int_a^b \frac {e^\frac bx}x dx} & \small \blue{\text{Let }u = \frac xa \implies du = - \frac {dx}a} \\ & = \blue{\int_1^\frac ba \frac {e^u}u du} - \red{\int_\frac ba^1 \frac {- e^v}v dv} & \small \red{\text{Let }v = \frac bx \implies dv = - \frac b{x^2}dx} \\ & = \int_1^\frac ba \frac {e^x}x dx - \int_1^\frac ba \frac {e^x}x dx & \small \blue{\text{Replace }u \text{ and }v \text{ with }x} \\ & = \boxed 0 \end{aligned}$