Answer
$$f\left( t \right) = - \cos t + {t^2} + 6$$
Work Step by Step
$$\eqalign{
& f'\left( t \right) = \sin t + 2t,{\text{ }}f\left( 0 \right) = 5 \cr
& f\left( t \right) = \int {f'\left( t \right)} dt \cr
& f\left( t \right) = \int {\left( {\sin t + 2t} \right)} dt \cr
& {\text{find the antiderivative}} \cr
& f\left( t \right) = - \cos t + {t^2} + C \cr
& \cr
& {\text{with }}f\left( 0 \right) = 5 \cr
& 5 = - \cos \left( 0 \right) + {\left( 0 \right)^2} + C \cr
& 5 = - 1 + C \cr
& C = 6 \cr
& then{\text{ }} \cr
& f\left( t \right) = - \cos t + {t^2} + 6 \cr} $$
