Answer
$2t^{(3/2)}-\dfrac{4}{t}+c$
Work Step by Step
Use formula $\int x^{a} dx=\dfrac{x^{(a+1)}}{(a+1)}+c$
where $c$ refers to a constant of proportionality.
Then, we have $\int (3 \sqrt t+\dfrac{4}{t^2}) dt=\int (3 t^{(1/2)}+4t^{-2}) dt$
and $(3)[\dfrac{t^{(\dfrac{1}{2})+1}}{\dfrac{1}{2}+1}]+ 4[\dfrac{t^{(-2+1)}}{(-2+1)}]+c=2t^{(3/2)}-\dfrac{4}{t}+c$
