Question Number 194434 by SANOGO last updated on 06/Jul/23
$${calcul} \\ $$$${e}^{\mathrm{2}{ln}\left(\mathrm{1}+{u}\right)\:} −{e}^{−\mathrm{2}{ln}\left(\mathrm{1}+{u}\right)} \:=? \\ $$
Answered by aba last updated on 06/Jul/23
$$\mathrm{e}^{\mathrm{2ln}\left(\mathrm{1}+\mathrm{u}\right)} −\mathrm{e}^{−\mathrm{2ln}\left(\mathrm{1}+\mathrm{u}\right)} =\mathrm{e}^{\mathrm{ln}\left(\mathrm{1}+\mathrm{u}\right)^{\mathrm{2}} } −\mathrm{e}^{\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{1}+\mathrm{u}}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{1}+\mathrm{u}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{u}\right)^{\mathrm{2}} } \\ $$