Question Number 25109 by ifcrna380w last updated on 04/Dec/17
Answered by mrW1 last updated on 04/Dec/17
$$\frac{{dy}}{{y}^{\mathrm{2}} }=\frac{{dx}}{{e}^{{x}} +{e}^{−{x}} }=\frac{\mathrm{1}}{\left({e}^{{x}} \right)^{\mathrm{2}} +\mathrm{1}}{d}\left({e}^{{x}} \right)=\frac{{du}}{{u}^{\mathrm{2}} +\mathrm{1}} \\ $$$$−\frac{\mathrm{1}}{{y}}=\mathrm{tan}^{−\mathrm{1}} {u}+{C}=\mathrm{tan}^{−\mathrm{1}} {e}^{{x}} +{C} \\ $$$$\Rightarrow{y}=−\frac{\mathrm{1}}{\mathrm{tan}^{−\mathrm{1}} {e}^{{x}} +{C}} \\ $$