Question Number 197362 by sonukgindia last updated on 14/Sep/23
Answered by MM42 last updated on 15/Sep/23
$$\frac{{e}^{{x}} −\mathrm{1}}{{e}^{{x}} +\mathrm{1}}={u}\Rightarrow{e}^{{x}} {u}+{u}={e}^{{x}} −\mathrm{1}\Rightarrow{e}^{{x}} \left({u}−\mathrm{1}\right)=−\mathrm{1}−{u} \\ $$$${e}^{{x}} =\frac{\mathrm{1}+{u}}{\mathrm{1}−{u}}\Rightarrow{f}\left({x}\right)=\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right)^{\mathrm{2}} \:\:\checkmark \\ $$
Answered by ibrahimabdullayev last updated on 15/Sep/23
$$\frac{{e}^{{x}} −\mathrm{1}}{{e}^{{x}} +\mathrm{1}}={a}\:\Rightarrow{e}^{{x}} −\mathrm{1}={ae}^{{x}} +{a}\:\Rightarrow{e}^{{x}} \left(\mathrm{1}−{a}\right)={a}+\mathrm{1} \\ $$$${e}^{{x}} =\frac{\mathrm{1}+{a}}{\mathrm{1}−{a}}\:\Rightarrow{f}\left({a}\right)=\left(\frac{\mathrm{1}+{a}}{\mathrm{1}−{a}}\right)^{\mathrm{2}} \Rightarrow{f}\left({x}\right)=\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:{Ibrahim}\:{Abdullayev} \\ $$