Answer
$$ - \frac{5}{t} + \frac{4}{3}{t^3} + C$$
Work Step by Step
$$\eqalign{
& \int {\left( {\frac{5}{{{t^2}}} + 4{t^2}} \right)} dt \cr
& {\text{use }}\frac{1}{{{t^n}}} = {t^{ - n}} \cr
& = \int {\left( {5{t^{ - 2}} + 4{t^2}} \right)} dt \cr
& {\text{use power rule for indefinite integrals}} \cr
& = \frac{{5{t^{ - 2 + 1}}}}{{ - 2 + 1}} + \frac{{4{t^{2 + 1}}}}{{2 + 1}} + C \cr
& = \frac{{5{t^{ - 1}}}}{{ - 1}} + \frac{{4{t^3}}}{3} + C \cr
& = - \frac{5}{t} + \frac{4}{3}{t^3} + C \cr
& \cr
& \cr
& {\text{check the antiderivative by differentiation}} \cr
& {\text{ = }}\frac{d}{{dt}}\left( { - 5{t^{ - 1}} + \frac{{4{t^3}}}{3} + C} \right) \cr
& {\text{ = }} - 5\left( { - 1} \right){t^{ - 2}} + \frac{{4\left( 3 \right){t^2}}}{3} + 0 \cr
& {\text{simplify}} \cr
& {\text{ = }}5{t^{ - 2}} + 4{t^2} \cr
& {\text{ = }}\frac{5}{{{t^2}}} + 4{t^2} \cr} $$
