Answer
$$\frac{{{2^t}{\bf{i}}}}{{\ln 2}} + \frac{1}{2}\ln \left| {1 + 2t} \right|{\bf{j}} + \left( { - t + t\ln t} \right){\bf{k}} + {\bf{C}}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = {2^t}{\bf{i}} + \frac{1}{{1 + 2t}}{\bf{j}} + \ln t{\bf{k}} \cr
& \int {{\bf{r}}\left( t \right)dt = } {\bf{i}}\int {{2^t}dt} + {\bf{j}}\int {\frac{1}{{1 + 2t}}dt} + {\bf{k}}\int {\ln t} dt \cr
& = \frac{1}{{\ln 2}}{\bf{i}}\int {{2^t}\left( {\ln 2} \right)dt} + \frac{1}{2}{\bf{j}}\int {\frac{2}{{1 + 2t}}dt} + {\bf{k}}\int {\ln t} dt \cr
& = \frac{{{2^t}{\bf{i}}}}{{\ln 2}} + \frac{1}{2}\ln \left| {1 + 2t} \right|{\bf{j}} + \left( { - t + t\ln t} \right){\bf{k}} + {\bf{C}} \cr
& {\text{Where }}{\bf{C}}{\text{ = }}{C_1}{\bf{i}} + {C_2}{\bf{j}} + {C_3}{\bf{k}} \cr} $$
