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: 56


$$ 2\sqrt{e^x+1}-2\sqrt{e^x+1} \coth^{-1}(\sqrt{e^x+1})+C$$

Work Step by Step

Given $$\int \sqrt{e^{x}+1} d x$$ Let $$ u^2 =e^x+1\ \ \ \ \ \ \ \ \ 2udu=e^xdx\to dx=\frac{2u}{u^2-1}du$$ Then \begin{align*} \int \sqrt{e^{x}+1} d x&=2\int \frac{u^2}{u^2-1}du\\ &=2\int \left(1+ \frac{1}{u^2-1}\right)du\\ &=2u-2u\begin{cases}\tanh^{-1}u&|u|<1\\ \coth^{-1}u&|u|>1\end{cases}+C\\ &=2\sqrt{e^x+1}-2\sqrt{e^x+1}\left(\begin{cases}\tanh^{-1}(\sqrt{e^x+1})&|\sqrt{e^x+1}|<1\\ \coth^{-1}(\sqrt{e^x+1})&|\sqrt{e^x+1}|>1\end{cases}+C\right) \end{align*} Since $e^x+1>1$, then $$ \int \sqrt{e^{x}+1} d x= 2\sqrt{e^x+1}-2\sqrt{e^x+1} \coth^{-1}(\sqrt{e^x+1})+C$$ Note that we can also write the result in log form, by converting $\coth^{-1}$ to $\ln$.
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