Answer
$$\frac{{52}}{3}{\bf{i}} + \frac{4}{3}{\bf{j}}$$
Work Step by Step
$$\eqalign{
& \int_1^9 {\left( {{t^{1/2}}{\bf{i}} + {t^{ - 1/2}}{\bf{j}}} \right)} dt \cr
& {\text{Integrating the components yields}} \cr
& \int_1^9 {\left( {{t^{1/2}}{\bf{i}} + {t^{ - 1/2}}{\bf{j}}} \right)} dt = \left[ {\frac{{{t^{3/2}}}}{{3/2}}} \right]_1^9{\bf{i}} + \left[ {\frac{{{t^{1/2}}}}{{1/2}}} \right]_1^9{\bf{j}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ {\frac{{2{t^{3/2}}}}{3}} \right]_1^9{\bf{i}} + \left[ {2{t^{1/2}}} \right]_1^9{\bf{j}} \cr
& {\text{Use the vector form of the Fundamental Theorem of Calculus}} \cr
& \int_a^b {{\bf{r}}\left( t \right)} dt = \left[ {{\bf{R}}\left( t \right)} \right]_a^b = {\bf{R}}\left( b \right) - {\bf{R}}\left( a \right) \cr
& {\text{then}} \cr
& \int_1^9 {\left( {{t^{1/2}}{\bf{i}} + {t^{ - 1/2}}{\bf{j}}} \right)} dt = \left[ {\frac{{2{{\left( 9 \right)}^{3/2}}}}{3} - \frac{{2{{\left( 1 \right)}^{3/2}}}}{3}} \right]{\bf{i}} + \left[ {\frac{{2{{\left( 9 \right)}^{1/2}}}}{3} - \frac{{2{{\left( 1 \right)}^{1/2}}}}{3}} \right]{\bf{j}} \cr
& {\text{simplify}} \cr
& \int_1^9 {\left( {{t^{1/2}}{\bf{i}} + {t^{ - 1/2}}{\bf{j}}} \right)} dt = \left[ {18 - \frac{2}{3}} \right]{\bf{i}} + \left[ {2 - \frac{2}{3}} \right]{\bf{j}} \cr
& \int_1^9 {\left( {{t^{1/2}}{\bf{i}} + {t^{ - 1/2}}{\bf{j}}} \right)} dt = \frac{{52}}{3}{\bf{i}} + \frac{4}{3}{\bf{j}} \cr} $$
