Answer
$2t^{3/2}-\dfrac{4}{t}+c$
Work Step by Step
Since, we know $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+c$
where $c$ is a constant of proportionality.
Now, $\int (3 \sqrt t+\dfrac{4}{t^2}) dt=\int (3 t^{1/2}+4t^{-2}) dt=\dfrac{3t^{\dfrac{1}{2}+1}}{\dfrac{1}{2}+1}+ \dfrac{4t^{-2+1}}{-2+1}+c$
or, $=2t^{3/2}-\dfrac{4}{t}+c$
