Answer
$\sqrt t+\dfrac{1}{t^3}+c$
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, $\int (\dfrac{1}{2} t^{-1/2}-3t^{-4}dt=\dfrac{1}{2}[\dfrac{t^{\dfrac{-1}{2}+1}}{\dfrac{-1}{2}+1}]+ \dfrac{3t^{-4+1}}{-4+1}+c$
or, $=t^{1/2}+\dfrac{1}{t^3}+c$
or, $=\sqrt t+\dfrac{1}{t^3}+c$
