Answer
$2 \sqrt t-\dfrac{2}{\sqrt t}+C$
Work Step by Step
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, $\int \dfrac{t\sqrt t+\sqrt t}{t^2} dt=\int \dfrac{t^{3/2}+t^{1/2}}{t^2} dt=\dfrac{1}{1/2}t^{1/2}+\dfrac{1}{-1/2}t^{-1/2}+C=2 \sqrt t-\dfrac{2}{\sqrt t}+C$
