Answer
$$s = \sqrt {21} + \frac{5}{4}\ln \left( {\frac{{4 + \sqrt {21} }}{{\sqrt 5 }}} \right)$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = t{\bf{i}} + {t^2}{\bf{j}} + 2t{\bf{k}},{\text{ }}\left[ {0,2} \right] \cr
& {\text{Differentiate }}{\bf{r}}\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {t{\bf{i}} + {t^2}{\bf{j}} + 2t{\bf{k}}} \right] \cr
& {\bf{r}}'\left( t \right) = {\bf{i}} + 2t{\bf{j}} + 2{\bf{k}} \cr
& {\text{Find }}\left\| {{\bf{r}}'\left( t \right)} \right\| \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {{{\left( 1 \right)}^2} + {{\left( {2t} \right)}^2} + {{\left( 2 \right)}^2}} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {4{t^2} + 5} \cr
& {\text{Find the arc length }}s{\text{, using the formula }}s = \int_a^b {\left\| {{\bf{r}}'\left( t \right)} \right\|} dt \cr
& s = \int_0^2 {\sqrt {4{t^2} + 5} } dt \cr
& s = \frac{1}{2}\int_0^2 {\sqrt {{{\left( {2t} \right)}^2} + 5} \left( 2 \right)} dt \cr
& {\text{Integrate by tables formula 26}}{\text{.}} \cr
& \int {\sqrt {{u^2} + {a^2}} du} = \frac{1}{2}\left( {u\sqrt {{u^2} + {a^2}} + {a^2}\ln \left| {u + \sqrt {{u^2} + {a^2}} } \right|} \right) + C \cr
& s = \frac{1}{4}\left[ {2t\sqrt {4{t^2} + 5} + 5\ln \left| {2t + \sqrt {4{t^2} + 5} } \right|} \right]_0^2 \cr
& s = \frac{1}{4}\left[ {2\left( 2 \right)\sqrt {4{{\left( 2 \right)}^2} + 5} + 5\ln \left| {2\left( 2 \right) + \sqrt {4{{\left( 2 \right)}^2} + 5} } \right|} \right] \cr
& - \frac{1}{4}\left[ {2\left( 0 \right)\sqrt {4{{\left( 0 \right)}^2} + 5} + 5\ln \left| {2\left( 0 \right) + \sqrt {4{{\left( 0 \right)}^2} + 5} } \right|} \right] \cr
& s = \frac{1}{4}\left[ {4\sqrt {21} + 5\ln \left| {4 + \sqrt {21} } \right|} \right] - \frac{1}{4}\left[ {5\ln \sqrt 5 } \right] \cr
& s = \frac{1}{4}\left[ {4\sqrt {21} + 5\ln \left| {4 + \sqrt {21} } \right| - 5\ln \sqrt 5 } \right] \cr
& s = \frac{1}{4}\left[ {4\sqrt {21} + 5\ln \left| {\frac{{4 + \sqrt {21} }}{{\sqrt 5 }}} \right|} \right] \cr
& s = \sqrt {21} + \frac{5}{4}\ln \left( {\frac{{4 + \sqrt {21} }}{{\sqrt 5 }}} \right) \cr
& {\text{Graph}} \cr} $$
