University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{(1.3)^x}{\ln (1.3)}+C$
Given: $\int (1.3)^x dx$ Use formula: $\int a^x dx= (\dfrac{1}{\ln a}) a^x +C$ Thus, $(1.3)^x dx=[\dfrac{1}{\ln (1.3)}] (1.3)^x +C=\dfrac{(1.3)^x}{\ln (1.3)}+C$