Answer
$$s = 3\sqrt {145} + \frac{1}{4}\ln \left( {12 + \sqrt {145} } \right)$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left( {t + 1} \right){\bf{i}} + {t^2}{\bf{j}},{\text{ }}\left[ {0,6} \right] \cr
& {\text{Differentiate }}{\bf{r}}\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {\left( {t + 1} \right){\bf{i}} + {t^2}{\bf{j}}} \right] \cr
& {\bf{r}}'\left( t \right) = {\bf{i}} + 2t{\bf{j}} \cr
& {\text{Find }}\left\| {{\bf{r}}'\left( t \right)} \right\| \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {{{\left( 1 \right)}^2} + {{\left( {2t} \right)}^2}} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {1 + 4{t^2}} \cr
& {\text{Find the arc length }}s{\text{ using the formula }}s = \int_a^b {\left\| {{\bf{r}}'\left( t \right)} \right\|} dt \cr
& s = \int_0^6 {\sqrt {1 + 4{t^2}} } dt \cr
& s = \frac{1}{2}\int_0^6 {\sqrt {{{\left( {2t} \right)}^2} + 1} \left( 2 \right)} dt \cr
& {\text{Integrate by tables using formula 26}}{\text{.}} \cr
& \int {\sqrt {{u^2} + {a^2}} du} = \frac{1}{2}\left( {u\sqrt {{u^2} + {a^2}} + {a^2}\ln \left| {u + \sqrt {{u^2} + {a^2}} } \right|} \right) + C \cr
& s = \frac{1}{4}\left[ {2t\sqrt {4{t^2} + 1} + \ln \left| {2t + \sqrt {4{t^2} + 1} } \right|} \right]_0^6 \cr
& s = \frac{1}{4}\left[ {2\left( 6 \right)\sqrt {4{{\left( 6 \right)}^2} + 1} + \ln \left| {2\left( 6 \right) + \sqrt {4{{\left( 6 \right)}^2} + 1} } \right|} \right] \cr
& - \frac{1}{4}\left[ {2\left( 0 \right)\sqrt {4{{\left( 0 \right)}^2} + 1} + \ln \left| {2\left( 0 \right) + \sqrt {4{{\left( 0 \right)}^2} + 1} } \right|} \right] \cr
& {\text{Simplifying}} \cr
& s = \frac{1}{4}\left[ {12\sqrt {145} + \ln \left| {12 + \sqrt {145} } \right|} \right] - \frac{1}{4}\left[ 0 \right] \cr
& s = 3\sqrt {145} + \frac{1}{4}\ln \left( {12 + \sqrt {145} } \right) \cr
& {\text{Graph}} \cr} $$
