Answer
$$\frac{1}{3}{t^3}{\bf{i}} - {t^2}{\bf{j}} + \ln \left| t \right|{\bf{k}} + {\bf{C}}$$
Work Step by Step
$$\eqalign{
& \int {\left( {{t^2}{\bf{i}} - 2t{\bf{j}} + \frac{1}{t}{\bf{k}}} \right)dt} \cr
& {\text{Integrating}} \cr
& {\bf{i}}\int {{t^2}} dt - {\bf{j}}\int {2t} dt + {\bf{k}}\int {\frac{1}{t}} dt \cr
& {\bf{i}}\left( {\frac{1}{3}{t^3} + {C_1}} \right) - {\bf{j}}\left( {{t^2} + {C_2}} \right) + \left( {\ln \left| t \right| + {C_3}} \right){\bf{k}} \cr
& {\text{Simplifying, }}{\bf{C}} = {C_1}{\bf{i}} + {C_2}{\bf{j}} + {C_3}{\bf{j}}{\text{ is an arbitrary vector constant }} \cr
& \frac{1}{3}{t^3}{\bf{i}} - {t^2}{\bf{j}} + \ln \left| t \right|{\bf{k}} + {\bf{C}} \cr} $$
