Answer
$${\bf{r}}'\left( t \right) = \left[ {\frac{1}{2}\ln \left( {{t^2} + 1} \right) + 1} \right]{\bf{i}} - \left( {\frac{1}{2}{e^{ - {t^2}}} - 2} \right){\bf{j}} - \left( {2\sqrt {t + 4} - 1} \right){\bf{k}}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}'\left( t \right) = \frac{t}{{{t^2} + 1}}{\bf{i}} + t{e^{ - {t^2}}}{\bf{j}} - \frac{{2t}}{{\sqrt {{t^2} + 4} }}{\bf{k}};\,\,\,\,\,{\bf{r}}\left( 0 \right) = {\bf{i}} + \frac{3}{2}{\bf{j}} - 3{\bf{k}} \cr
& {\bf{r}}'\left( t \right) = {\bf{i}}\int {\frac{t}{{{t^2} + 1}}} dt + {\bf{j}}\int {t{e^{ - {t^2}}}dt} - {\bf{k}}\int {\frac{{2t}}{{\sqrt {{t^2} + 4} }}} dt \cr
& {\bf{r}}'\left( t \right) = \frac{1}{2}{\bf{i}}\int {\frac{{2t}}{{{t^2} + 1}}} dt + \frac{1}{{ - 2}}{\bf{j}}\int {\left( { - 2t} \right){e^{ - {t^2}}}dt} - {\bf{k}}\int {2t{{\left( {{t^2} + 4} \right)}^{ - 1/2}}} dt \cr
& {\bf{r}}'\left( t \right) = \frac{1}{2}\ln \left( {{t^2} + 1} \right){\bf{i}} - \frac{1}{2}{e^{ - {t^2}}}{\bf{j}} - 2\sqrt {t + 4} {\bf{k}} + {\bf{C}}\,\,\,\left( 1 \right) \cr
& \cr
& {\text{Use the initial condition}} \cr
& {\bf{i}} + \frac{3}{2}{\bf{j}} - 3{\bf{k}} = \frac{1}{2}\ln \left( {{0^2} + 1} \right){\bf{i}} - \frac{1}{2}{e^{ - {0^2}}}{\bf{j}} - 2\sqrt {0 + 4} {\bf{k}} + {\bf{C}} \cr
& {\bf{i}} + \frac{3}{2}{\bf{j}} - 3{\bf{k}} = 0{\bf{i}} - \frac{1}{2}{\bf{j}} - 4{\bf{k}} + {\bf{C}} \cr
& {\bf{C}} = {\bf{i}} + 2{\bf{j}} + {\bf{k}} \cr
& \cr
& {\text{Substitute the vector constant into }}\left( 1 \right) \cr
& {\bf{r}}'\left( t \right) = \frac{1}{2}\ln \left( {{t^2} + 1} \right){\bf{i}} - \frac{1}{2}{e^{ - {t^2}}}{\bf{j}} - 2\sqrt {t + 4} {\bf{k}} + {\bf{i}} + \frac{3}{2}{\bf{j}} - 3{\bf{k}} + \frac{1}{2}{\bf{j}} + 4{\bf{k}} \cr
& {\bf{r}}'\left( t \right) = \left[ {\frac{1}{2}\ln \left( {{t^2} + 1} \right) + 1} \right]{\bf{i}} - \left( {\frac{1}{2}{e^{ - {t^2}}} - 2} \right){\bf{j}} - \left( {2\sqrt {t + 4} - 1} \right){\bf{k}} \cr} $$
