Answer
$$ - \frac{1}{t} - \frac{1}{{{t^2}}} + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{{{\left( {t + 1} \right)}^2} - 1}}{{{t^4}}}} dt \cr
& {\text{expanding }}{\left( {t + 1} \right)^2} \text{ gives} \cr
& = \int {\frac{{{t^2} + 2t + 1 - 1}}{{{t^4}}}} dt \cr
& = \int {\frac{{{t^2} + 2t}}{{{t^4}}}} dt \cr
& {\text{distribute and simplify}} \cr
& = \int {\left( {\frac{{{t^2}}}{{{t^4}}} + \frac{{2t}}{{{t^4}}}} \right)} dt \cr
& = \int {\left( {{t^{ - 2}} + 2{t^{ - 3}}} \right)} dt \cr
& {\text{integrate by using the power rule}} \cr
& = \frac{{{t^{ - 1}}}}{{ - 1}} + 2\left( {\frac{{{t^{ - 2}}}}{{ - 2}}} \right) + C \cr
& = - \frac{1}{t} - \frac{1}{{{t^2}}} + C \cr} $$
