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Question Number 132473 by physicstutes last updated on 14/Feb/21

$$\int \frac{x\cosh x}{{\left(\sinh x\right)}^2}dx$$

Answered by mathmax by abdo last updated on 14/Feb/21

$$\mathrm{I}=\int \frac{\mathrm{xchx}}{{\mathrm{sh}}^2\mathrm{x}}\mathrm{dx}\mathrm{by}\mathrm{parts}{\mathrm{u}}^{{}^{\prime }}=\frac{\mathrm{chx}}{{\mathrm{sh}}^2\mathrm{x}}\mathrm{and}\mathrm{v}=\mathrm{x}\Rightarrow$$$$\mathrm{I}=-\frac{\mathrm{x}}{\mathrm{shx}}-\int (-\frac{1}{\mathrm{shx}})\mathrm{dx}=-\frac{\mathrm{x}}{\mathrm{shx}}+\int \frac{\mathrm{dx}}{\mathrm{shx}}\mathrm{we}\mathrm{have}$$$$\int \frac{\mathrm{dx}}{\mathrm{shx}}=2\int \frac{\mathrm{dx}}{{\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{-\mathrm{x}}}{=}_{{\mathrm{e}}^{\mathrm{x}}=\mathrm{t}}2\int \frac{\mathrm{dt}}{\mathrm{t}(\mathrm{t}-{\mathrm{t}}^{-1})}=\int \frac{2\mathrm{dt}}{{\mathrm{t}}^2-1}$$$$=\int (\frac{1}{\mathrm{t}-1}-\frac{1}{\mathrm{t}+1})\mathrm{dt}=\ln \mid \frac{\mathrm{t}-1}{\mathrm{t}+1}\mid +\mathrm{C}=\ln \mid \frac{{\mathrm{e}}^{\mathrm{x}}-1}{{\mathrm{e}}^{\mathrm{x}}+1}\mid +\mathrm{C}\Rightarrow$$$$\mathrm{I}=-\frac{\mathrm{x}}{\mathrm{shx}}+\ln \mid \frac{{\mathrm{e}}^{\mathrm{x}}-1}{{\mathrm{e}}^{\mathrm{x}}+1}\mid +\mathrm{C}$$

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