## Calculus: Early Transcendentals (2nd Edition)

$$\frac{1}{3}\ln \left( {\frac{{65}}{{16}}} \right)$$
\eqalign{ & \int_0^{\ln 2} {\frac{{{e^{3x}} - {e^{ - 3x}}}}{{{e^{3x}} + {e^{ - 3x}}}}} dx \cr & {\text{substitute }}u = {e^{3x}} + {e^{ - 3x}},{\text{ }}du = 3{e^{3x}} - 3{e^{ - 3x}}dx \cr & {\text{express the limits in terms of }}u \cr & x = \ln 2{\text{ implies }}u = {e^{3\left( {\ln 2} \right)}} + {e^{ - 3\left( {\ln 2} \right)}} = 65/8 \cr & x = 0{\text{ implies }}u = {e^{3x}} + {e^{ - 3x}} = 2 \cr & {\text{the entire integration is carried out as follows}} \cr & \int_0^{\ln 2} {\frac{{{e^{3x}} - {e^{ - 3x}}}}{{{e^{3x}} + {e^{ - 3x}}}}} dx = \frac{1}{3}\int_2^{65/8} {\frac{{du}}{u}} \cr & {\text{find the antiderivative}} \cr & = \left. {\frac{1}{3}\left( {\ln \left| u \right|} \right)} \right|_2^{65/8} \cr & {\text{evaluate limits}} \cr & = \frac{1}{3}\left( {\ln \left( {\frac{{65}}{8}} \right) - \ln \left( 2 \right)} \right) \cr & {\text{simplify}} \cr & = \frac{1}{3}\ln \left( {\frac{{65}}{{16}}} \right) \cr}