Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - 8.6 Strategies for Integration - Exercises - Page 431: 48


$$\frac{1}{2} e^x \sqrt{e^{2x}-1} -\frac{1}{2} \ln |e^x +\sqrt{e^{2x}-1}|+C$$

Work Step by Step

Given $$\int e^{x} \sqrt{e^{2 x}-1} d x$$ Let $$e^x= \sec u\ \ \ \ \ \ \ \ e^xdx=\sec u\tan udu $$ Then \begin{align*} \int e^{x} \sqrt{e^{2 x}-1} d x&=\int \sqrt{\sec^2 u-1} \sec u\tan udu \\ &=\int \tan^2 u \sec udu\\ &=\int (\sec^2 u-1)\sec udu\\ &=\int (\sec^3 u-\sec u)du \end{align*} Use $$\int \sec ^{3} u d u=\frac{1}{2} \sec u \tan u+\frac{1}{2} \ln |\sec u+\tan u|+C $$ Then \begin{align*} \int e^{x} \sqrt{e^{2 x}-1} d x &=\int (\sec^3 u-\sec u)du\\ &=\frac{1}{2} \sec u \tan u-\frac{1}{2} \ln |\sec u+\tan u|+C \\ &=\frac{1}{2} e^x \sqrt{e^{2x}-1} -\frac{1}{2} \ln |e^x +\sqrt{e^{2x}-1}|+C \end{align*}
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