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Question Number 73258 by Lontum Hans last updated on 09/Nov/19

Answered by MJS last updated on 09/Nov/19

$$\mathrm{well},\mathrm{just}\mathrm{do}\mathrm{it}$$$$u=1+\cosh x\to dx=\frac{du}{\sinh x}$$$$...=\underset{2}{\stackrel{3}{\int }}\frac{du}{u(u-1)}={\left[\ln \frac{u-1}{u}\right]}_2^3=\ln \frac{4}{3}$$

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