Answer
$f(t) = t^2+3~cos~t+2$
Work Step by Step
$f'(t) = 2t-3~sin~t$
$f(t) = \int (2t-3~sin~t)~dt = t^2+3~cos~t+C$
Note that $f(0) = 5$
We can find the value of $C$:
$f(0) = (0)^2+3~cos~(0)+C = 5$
$0+3+C = 5$
$C = 2$
Therefore:
$f(t) = t^2+3~cos~t+2$