Answer
$$L = 4\sqrt 5 $$
Work Step by Step
$$\eqalign{
& y = 2x + 1{\text{ on }}\left[ {1,5} \right] \cr
& {\text{Definition of Arc Length for }}y = f\left( x \right): \cr
& {\text{Let }}f{\text{ have a continuous first derivative on the interval }}\left[ {a,b} \right]{\text{ The length of the curve }} \cr
& {\text{from }}\left( {a,f\left( a \right)} \right){\text{ to }}\left( {b,f\left( b \right)} \right){\text{ is }}L = \int_a^b {\sqrt {1 + f'{{\left( x \right)}^2}} } dx \cr
& {\text{Notice that }}y = f\left( x \right) = 2x + 1{\text{ and }}\left[ {1,5} \right] \to a = 1{\text{ and }}b = 5.{\text{ then}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {2x + 1} \right] \cr
& f'\left( x \right) = 2 \cr
& {\text{Using the arc length formula}}{\text{, we have}} \cr
& L = \int_1^5 {\sqrt {1 + {{\left( 2 \right)}^2}} } dx \cr
& {\text{simplifying}} \cr
& L = \int_1^5 {\sqrt 5 } dx \cr
& {\text{integrate}} \cr
& L = \sqrt 5 \left( x \right)_1^5 \cr
& L = \sqrt 5 \left( {5 - 1} \right) \cr
& L = 4\sqrt 5 \cr} $$
