Answer
$$\frac{{{t^3}}}{3} + \frac{4}{t} + C $$
Work Step by Step
$$\eqalign{
& \int {\left( {t - \frac{2}{t}} \right)\left( {t + \frac{2}{t}} \right)} dt \cr
& {\text{multiply the expression }}\left( {t - \frac{2}{t}} \right)\left( {t + \frac{2}{t}} \right) \cr
& \left( {t - \frac{2}{t}} \right)\left( {t + \frac{2}{t}} \right) = {\left( t \right)^2} - {\left( {\frac{2}{t}} \right)^2} \cr
& \left( {t - \frac{2}{t}} \right)\left( {t + \frac{2}{t}} \right) = {t^2} - \frac{4}{{{t^2}}} \cr
& \left( {t - \frac{2}{t}} \right)\left( {t + \frac{2}{t}} \right) = {t^2} - 4{t^{ - 2}} \cr
& {\text{Then}} \cr
& \int {\left( {t - \frac{2}{t}} \right)\left( {t + \frac{2}{t}} \right)} dt = \int {\left( {{t^2} - 4{t^{ - 2}}} \right)} dt \cr
& {\text{integrate by using the power rule}} \cr
& = \frac{{{t^3}}}{3} - 4\left( {\frac{{{t^{ - 1}}}}{{ - 1}}} \right) + C \cr
& {\text{simplifying, we get:}} \cr
& = \frac{{{t^3}}}{3} + \frac{4}{t} + C \cr} $$
