Question Number 196596 by MATHEMATICSAM last updated on 27/Aug/23
$$\mathrm{If}\:{f}\left({x}\right)\:=\:\mathrm{ln}\left(\frac{\mathrm{1}\:+\:{x}}{\mathrm{1}\:−\:{x}}\right)\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$${f}\left(\frac{\mathrm{2}{x}}{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\right)\:=\:\mathrm{2}{f}\left({x}\right). \\ $$
Commented by mokys last updated on 27/Aug/23
$${f}\:\left(\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\right)=\:{ln}\left(\frac{\mathrm{1}+\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }}{\mathrm{1}−\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }}\:\right)\:=\:{ln}\:\left(\:\frac{\mathrm{1}+\mathrm{2}{x}+{x}^{\mathrm{2}} }{\mathrm{1}−\mathrm{2}{x}+{x}^{\mathrm{2}} }\right)\:\:\: \\ $$$$ \\ $$$$\:=\:{ln}\:\left(\:\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right)^{\mathrm{2}} \:=\:\mathrm{2}\:{ln}\:\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right)\:=\:\mathrm{2}\:{f}\left({x}\right) \\ $$