Answer
$${\bf{r}}\left( t \right) = \left( {{t^2} + 1} \right){\bf{i}} + \left( {{e^t} + 2} \right){\bf{j}} - \left( {{e^{ - t}} + 4} \right){\bf{k}}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}'\left( t \right) = 2t{\bf{i}} + {e^t}{\bf{j}} + {e^{ - t}}{\bf{k}} \cr
& {\text{Integrating}} \cr
& {\bf{r}}\left( t \right) = \int {{\bf{r}}'\left( t \right)} dt \cr
& = \left[ {\int {2tdt} } \right]{\bf{i}} + \left[ {\int {{e^t}dt} } \right]{\bf{j}} + \left[ {\int {{e^{ - t}}dt} } \right]{\bf{k}} \cr
& {\bf{r}}\left( t \right) = {t^2}{\bf{i}} + {e^t}{\bf{j}} - {e^{ - t}}{\bf{k}} + {\bf{C}} \cr
& {\text{Using the fact that }}{\bf{r}}\left( 0 \right) = {\bf{i}} + 3{\bf{j}} - 5{\bf{k}} \cr
& {\bf{r}}\left( 0 \right) = {\left( 0 \right)^2}{\bf{i}} + {e^0}{\bf{j}} - {e^{ - 0}}{\bf{k}} + {\bf{C}} \cr
& {\bf{r}}\left( 0 \right) = 0{\bf{i}} + {\bf{j}} - {\bf{k}} + {\bf{C}} \cr
& {\bf{i}} + 3{\bf{j}} - 5{\bf{k}} = + {\bf{j}} - {\bf{k}} + {\bf{C}} \cr
& {\bf{C}} = {\bf{i}} + 3{\bf{j}} - 5{\bf{k}} - {\bf{j}} + {\bf{k}} \cr
& {\bf{C}} = {\bf{i}} + 2{\bf{j}} - 4{\bf{k}} \cr
& {\text{Therefore,}} \cr
& {\bf{r}}\left( t \right) = {t^2}{\bf{i}} + {e^t}{\bf{j}} - {e^{ - t}}{\bf{k}} + {\bf{i}} + 2{\bf{j}} - 4{\bf{k}} \cr
& {\bf{r}}\left( t \right) = \left( {{t^2} + 1} \right){\bf{i}} + \left( {{e^t} + 2} \right){\bf{j}} - \left( {{e^{ - t}} + 4} \right){\bf{k}} \cr} $$
