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