## Calculus, 10th Edition (Anton)

Published by Wiley

# Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.8 Hyperbolic Functions And Hanging Cables - Exercises Set 6.8 - Page 481: 39

#### Answer

$$- {\operatorname{sech} ^{ - 1}}{e^x} + C$$

#### Work Step by Step

\eqalign{ & \int {\frac{{dx}}{{\sqrt {1 - {e^{2x}}} }}} \cr & = \int {\frac{{dx}}{{\sqrt {1 - {{\left( {{e^x}} \right)}^2}} }}} \cr & {\text{substitute }}u = {e^x},{\text{ }}du = {e^x}dx \cr & = \int {\frac{{dx}}{{\sqrt {1 - {{\left( {{e^x}} \right)}^2}} }}} = \int {\frac{{du/{e^x}}}{{\sqrt {1 - {u^2}} }}} \cr & = \int {\frac{{du}}{{u\sqrt {1 - {u^2}} }}} \cr & {\text{find the antiderivarive using the theorem 6}}{\text{.8}}{\text{.6 }}\left( {{\text{see page 480}}} \right) \cr & = - {\operatorname{sech} ^{ - 1}}\left| u \right| + C \cr & {\text{write in terms of }}x \cr & = - {\operatorname{sech} ^{ - 1}}\left| {{e^x}} \right| + C \cr & = - {\operatorname{sech} ^{ - 1}}{e^x} + C \cr}

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