If $\left\{f\left(x\right)={\log }_e\left(\frac{1-x}{1+x}\right),\left|x\right|<1\right\}$ then $f\left(\frac{2x}{1+x^2}\right)$ is equal to,

Solution:

Here $f\left(x\right)={\log }_e\left(\frac{1-x}{1+x}\right)$

Now, $f\left(\frac{2x}{1+x^2}\right)={\log }_e\left(\frac{1-\frac{2x}{1+x^2}}{1+\frac{2x}{1+x^2}}\right)$

${\log }_e\left(\frac{1+x^2-2x}{1+x^2+2x}\right)={\log }_e{\left(\frac{1-x}{1+x}\right)}^2=2{\log }_e\left(\frac{1-x}{1+x}\right)=2f\left(x\right)$