Answer
$$ - \frac{1}{{2{t^2}}} - 2t + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{1 - 2{t^3}}}{{{t^3}}}} dt \cr
& {\text{distribute numerator}} \cr
& = \int {\left( {\frac{1}{{{t^3}}} - \frac{{2{t^3}}}{{{t^3}}}} \right)} dx \cr
& = \int {\left( {\frac{1}{{{t^3}}} - 2} \right)} dx \cr
& = \int {\left( {{t^{ - 3}} - 2} \right)} dx \cr
& {\text{power rule}} \cr
& = \frac{{{t^{ - 3 + 1}}}}{{ - 3 + 1}} - 2t + C \cr
& = \frac{{{t^{ - 2}}}}{{ - 2}} - 2t + C \cr
& = - \frac{1}{{2{t^2}}} - 2t + C \cr
& {\text{check by differentiation}} \cr
& = \frac{d}{{dx}}\left[ {\frac{{{t^{ - 2}}}}{{ - 2}} - 2t + C} \right] \cr
& = \frac{{ - 2{t^{ - 3}}}}{{ - 2}} - 2 + 0 \cr
& = \frac{1}{{{t^3}}} - 2 \cr} $$
