#### Answer

$$g\left( t \right) = \frac{{{t^3}}}{3} - \frac{1}{t} + \frac{5}{3}$$

#### Work Step by Step

$$\eqalign{
& g'\left( t \right) = {t^2} + {t^{ - 2}},{\text{ }}g\left( 1 \right) = 1 \cr
& g\left( t \right) = \int {g'\left( t \right)} dt \cr
& g\left( t \right) = \int {\left( {{t^2} + {t^{ - 2}}} \right)} dt \cr
& {\text{find the antiderivative by the power rule}} \cr
& g\left( t \right) = \frac{{{t^{2 + 1}}}}{{2 + 1}} + \frac{{{t^{ - 2 + 1}}}}{{ - 2 + 1}} + C \cr
& g\left( t \right) = \frac{{{t^3}}}{3} + \frac{{{t^{ - 1}}}}{{ - 1}} + C \cr
& g\left( t \right) = \frac{{{t^3}}}{3} - \frac{1}{t} + C \cr
& \cr
& {\text{with }}g\left( 1 \right) = 1 \cr
& 1 = \frac{{{{\left( 1 \right)}^3}}}{3} - \frac{1}{1} + C \cr
& 1 = - \frac{2}{3} + C \cr
& C = \frac{5}{3} \cr
& then{\text{ }} \cr
& g\left( t \right) = \frac{{{t^3}}}{3} - \frac{1}{t} + \frac{5}{3} \cr} $$