Question Number 73258 by Lontum Hans last updated on 09/Nov/19
Answered by MJS last updated on 09/Nov/19
![well, just do it u=1+cosh x → dx=(du/(sinh x)) ...=∫_2 ^3 (du/(u(u−1)))=[ln ((u−1)/u)]_2 ^3 =ln (4/3)](Q73267.png)
$$\mathrm{well},\:\mathrm{just}\:\mathrm{do}\:\mathrm{it} \\ $$$${u}=\mathrm{1}+\mathrm{cosh}\:{x}\:\rightarrow\:{dx}=\frac{{du}}{\mathrm{sinh}\:{x}} \\ $$$$...=\underset{\mathrm{2}} {\overset{\mathrm{3}} {\int}}\frac{{du}}{{u}\left({u}−\mathrm{1}\right)}=\left[\mathrm{ln}\:\frac{{u}−\mathrm{1}}{{u}}\right]_{\mathrm{2}} ^{\mathrm{3}} =\mathrm{ln}\:\frac{\mathrm{4}}{\mathrm{3}} \\ $$