Answer
$2t^4+\dfrac{t^3}{6}+\dfrac{t^2}{2}+c$
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, $\int (8t^3-\dfrac{t^2}{2}+t) dt=\dfrac{8t^{3+1}}{3+1}- \dfrac{t^{2+1}}{2+1}+\dfrac{t^2}{2}+c$
or, $=\dfrac{8t^4}{4}+\dfrac{t^3}{6}+\dfrac{t^2}{2}+c$
or, $=2t^4+\dfrac{t^3}{6}+\dfrac{t^2}{2}+c$
