Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.4 Exercises - Page 540: 61

$s = \ln \left( {\frac{{5\left( {1 + \sqrt 2 } \right)}}{{1 + \sqrt {26} }}} \right) + \sqrt {26} - \sqrt 2 $

$$\eqalign{ & y = \ln x{\text{ on the interval }}\left[ {1,5} \right] \cr & {\text{Differentiate}} \cr & \frac{{dy}}{{dx}} = \frac{1}{x} \cr & \cr & {\text{Use the arc length formula}} \cr & s = \int_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} } dx \cr & s = \int_1^5 {\sqrt {1 + {{\left( {\frac{1}{x}} \right)}^2}} } dx \cr & s = \int_1^5 {\sqrt {\frac{{{x^2} + 1}}{{{x^2}}}} } dx \cr & s = \int_1^5 {\frac{{\sqrt {{x^2} + 1} }}{x}} dx \cr & {\text{Integrate by substitution}}{\text{, from the triangle shown below}}{\text{,}} \cr & x = \tan \theta ,{\text{ }}dx = {\sec ^2}\theta d\theta \cr & \sqrt {{x^2} + 1} = \sec \theta \cr & {\text{Substituting}} \cr & s = \int_{}^{} {\frac{{\sqrt {{x^2} + 1} }}{x}} dx = \int_{}^{} {\frac{{\sec \theta }}{{\tan \theta }}\left( {{{\sec }^2}\theta } \right)} d\theta \cr & s = \int_{}^{} {\frac{{\sec \theta }}{{\tan \theta }}\left( {1 + {{\tan }^2}\theta } \right)} d\theta \cr & s = \int_{}^{} {\left( {\frac{{\sec \theta }}{{\tan \theta }} + \sec \theta \tan \theta } \right)} d\theta \cr & s = \int_{}^{} {\left( {\csc \theta + \sec \theta \tan \theta } \right)} d\theta \cr & {\text{Integrating}} \cr & s = - \ln \left| {\csc \theta + \cot \theta } \right| + \sec \theta + C \cr & {\text{From the triangle}} \cr & s = - \ln \left| {\frac{{\sqrt {{x^2} + 1} }}{x} + \frac{1}{x}} \right| + \sqrt {{x^2} + 1} + C \cr & s = - \ln \left| {\frac{{1 + \sqrt {{x^2} + 1} }}{x}} \right| + \sqrt {{x^2} + 1} + C \cr & {\text{Therefore}} \cr & s = \left[ { - \ln \left| {\frac{{1 + \sqrt {{x^2} + 1} }}{x}} \right| + \sqrt {{x^2} + 1} } \right]_1^5 + C \cr & {\text{Evaluating}} \cr & s = \left[ { - \ln \left| {\frac{{1 + \sqrt {25 + 1} }}{5}} \right| + \sqrt {25 + 1} } \right] - \left[ { - \ln \left| {\frac{{1 + \sqrt 2 }}{1}} \right| + \sqrt 2 } \right] \cr & {\text{Simplifying}} \cr & s = \left[ { - \ln \left( {\frac{{1 + \sqrt {26} }}{5}} \right) + \sqrt {26} } \right] - \left[ { - \ln \left( {1 + \sqrt 2 } \right) + \sqrt 2 } \right] \cr & s = - \ln \left( {\frac{{1 + \sqrt {26} }}{5}} \right) + \sqrt {26} + \ln \left( {1 + \sqrt 2 } \right) - \sqrt 2 \cr & s = \ln \left( {\frac{{5\left( {1 + \sqrt 2 } \right)}}{{1 + \sqrt {26} }}} \right) + \sqrt {26} - \sqrt 2 \approx 4.367 \cr} $$