## University Calculus: Early Transcendentals (3rd Edition)

$2 \sqrt t-\dfrac{2}{\sqrt t}+C$
Given: $\int \dfrac{t\sqrt t+\sqrt t}{t^2} dt$ Since, we have $\int x^n dx=\dfrac{x^{n+1}}{n+1}+C$ Thus, $\int \dfrac{t\sqrt t+\sqrt t}{t^2} dt=\int \dfrac{t^{3/2}+t^{1/2}}{t^2} dt=\dfrac{1}{1/2}t^{1/2}+\dfrac{1}{-1/2}t^{-1/2}+C=2 \sqrt t-\dfrac{2}{\sqrt t}+C$