Chapter 7 - Applications of Integration - 7.4 Exercises - Page 473: 14

$$s = \ln \left( {\frac{{16}}{9}} \right)$$

$$\eqalign{ & y = \ln \left( {\frac{{{e^x} + 1}}{{{e^x} - 1}}} \right),{\text{ }}\left[ {\ln 2,\ln 3} \right] \cr & {\text{Using logarithmic properties}} \cr & y = \ln \left( {{e^x} + 1} \right) - \ln \left( {{e^x} - 1} \right) \cr & {\text{Differentiate}} \cr & \frac{{dy}}{{dx}} = \frac{{{e^x}}}{{{e^x} + 1}} - \frac{{{e^x}}}{{{e^x} - 1}} \cr & \frac{{dy}}{{dx}} = \frac{{{e^{2x}} - {e^x} - {e^{2x}} - {e^x}}}{{{e^{2x}} - 1}} \cr & \frac{{dy}}{{dx}} = - \frac{{2{e^x}}}{{{e^{2x}} - 1}} \cr & {\text{Use the formula for arc length}} \cr & s = \int_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} } dx \cr & s = \int_{\ln 2}^{\ln 3} {\sqrt {1 + {{\left( { - \frac{{2{e^x}}}{{{e^{2x}} - 1}}} \right)}^2}} } dx \cr & s = \int_{\ln 2}^{\ln 3} {\sqrt {1 + \frac{{4{e^{2x}}}}{{{{\left( {{e^{2x}} - 1} \right)}^2}}}} } dx \cr & s = \int_{\ln 2}^{\ln 3} {\sqrt {\frac{{{{\left( {{e^{2x}} - 1} \right)}^2} + 4{e^{2x}}}}{{{{\left( {{e^{2x}} - 1} \right)}^2}}}} } dx \cr & s = \int_{\ln 2}^{\ln 3} {\sqrt {\frac{{{e^{4x}} - 2{e^{2x}} + 1 + 4{e^{2x}}}}{{{{\left( {{e^{2x}} - 1} \right)}^2}}}} } dx \cr & s = \int_{\ln 2}^{\ln 3} {\sqrt {\frac{{{e^{4x}} + 2{e^{2x}} + 1}}{{{{\left( {{e^{2x}} - 1} \right)}^2}}}} } dx \cr & s = \int_{\ln 2}^{\ln 3} {\sqrt {\frac{{{{\left( {{e^{2x}} + 1} \right)}^2}}}{{{{\left( {{e^{2x}} - 1} \right)}^2}}}} } dx \cr & s = \int_{\ln 2}^{\ln 3} {\frac{{{e^{2x}} + 1}}{{{e^{2x}} - 1}}} dx \cr & {\text{Split and integrate}} \cr & s = \int_{\ln 2}^{\ln 3} {\left( {\frac{{{e^{2x}}}}{{{e^{2x}} - 1}} + \frac{1}{{{e^{2x}} - 1}}} \right)} dx \cr & s = \int_{\ln 2}^{\ln 3} {\left( {\frac{{2{e^{2x}}}}{{2\left( {{e^{2x}} - 1} \right)}} + \frac{1}{{{e^{2x}} - 1}}} \right)} dx \cr & s = \left[ {\frac{1}{2}\ln \left| {{e^{2x}} - 1} \right| + \frac{1}{2}\ln \left| {1 - {e^{2x}}} \right| - x} \right]_{\ln 2}^{\ln 3} \cr & s = \left[ {\frac{1}{2}\ln \left| {{e^{2\left( {\ln 3} \right)}} - 1} \right| + \frac{1}{2}\ln \left| {1 - {e^{2\left( {\ln 3} \right)}}} \right| - \left( {\ln 3} \right)} \right] \cr & - \left[ {\frac{1}{2}\ln \left| {{e^{2\left( {\ln 2} \right)}} - 1} \right| + \frac{1}{2}\ln \left| {1 - {e^{2\left( {\ln 2} \right)}}} \right| - \left( {\ln 2} \right)} \right] \cr & s = \left[ {\frac{1}{2}\ln \left| {9 - 1} \right| + \frac{1}{2}\ln \left| {1 - 9} \right| - \left( {\ln 3} \right)} \right] \cr & - \left[ {\frac{1}{2}\ln \left| {4 - 1} \right| + \frac{1}{2}\ln \left| {1 - 4} \right| - \left( {\ln 2} \right)} \right] \cr & s = \left( {\ln 8 - \ln 3} \right) - \left( {\ln 3 - \ln 2} \right) \cr & s = \ln 8 - \ln 3 - \ln 3 + \ln 2 \cr & s = \ln 16 - 2\ln 3 \cr & s = \ln \left( {\frac{{16}}{9}} \right) \cr} $$