Answer
$$2{t^6} + \frac{1}{t} + C$$
Work Step by Step
$$\eqalign{
& {\text{split the numerator}} \cr
& = \int {\left( {\frac{{12{t^8}}}{{{t^3}}} - \frac{t}{{{t^3}}}} \right)dt} \cr
& = \int {\left( {12{t^5} - {t^{ - 2}}} \right)dt} \cr
& {\text{use power rule for indefinite integrals}} \cr
& = 12\left( {\frac{{{t^6}}}{6}} \right) - \left( {\frac{{{t^{ - 1}}}}{{ - 1}}} \right) + C \cr
& = 2{t^6} + \frac{1}{t} + C \cr
& {\text{check by differentiation}} \cr
& {\text{ = }}\frac{d}{{dt}}\left( {2{t^6} + \frac{1}{t} + C} \right) \cr
& = 12{t^5} - \frac{1}{{{t^2}}} + 0 \cr
& = \frac{{12{t^7} - 1}}{{{t^2}}} \cr
& = \frac{{12{t^8} - t}}{{{t^3}}} \cr} $$
