## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 6 - Applications of Integration - 6.10 Hyperbolic Functions - 6.10 Exercises - Page 503: 55

#### Answer

$$\frac{1}{6}{\tanh ^{ - 1}}\frac{{{e^x}}}{6} + C$$

#### Work Step by Step

\eqalign{ & \int {\frac{{{e^x}}}{{36 - {e^{2x}}}}dx} \cr & {\text{rewriting the denominator}} \cr & \int {\frac{{{e^x}}}{{{{\left( 6 \right)}^2} - {{\left( {{e^x}} \right)}^2}}}dx} \cr & {\text{find the antiderivative using the theorem 6}}{\text{.12}}{\text{, }}\left| {{e^x}} \right|{\text{ > 0}} \cr & \int {\frac{{dx}}{{{a^2} - {u^2}}}} = \frac{1}{a}{\tanh ^{ - 1}}\frac{u}{a} + C \cr & ,so \cr & = \frac{1}{6}{\tanh ^{ - 1}}\frac{{{e^x}}}{6} + C \cr}

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