Function exercise asked by Salma Al in many Arabic math groups

Exercise:

Find $f\left( x \right)$ such that$f\left( \frac{x-3}{x+1} \right)+f\left( \frac{x+3}{1-x} \right)=x$,  $x\in \mathbb{R}\,\And \,\,\left| x \right|\ne 1$

Solution: Let $u=\frac{x-3}{x+1}\Leftrightarrow ux+u=x-3\Leftrightarrow ux-x=-3-u\Leftrightarrow x\left( u-1 \right)=-3-u$

$\Leftrightarrow x=\frac{-3-u}{u-1}=\frac{-\left( 3+u \right)}{u-1}=\frac{3+u}{1-u}$

but $\frac{x+3}{1-x}=\frac{\frac{3+u}{1-u}+3}{1-\frac{3+u}{1-u}}=\frac{\frac{3+u+3-3u}{1-u}}{\frac{1-u-3-u}{1-u}}=\frac{6-2u}{-2u-2}=\frac{2\left( 3-u \right)}{-2\left( u+1 \right)}=\frac{u-3}{u+1}$

so $f\left( u \right)+f\left( \frac{u-3}{u+1} \right)=\frac{3+u}{1-u}$        (*)

Let $t=\frac{x+3}{1-x}\Leftrightarrow t-tx=x+3\Leftrightarrow x\left( -t-1 \right)=-t+3\Leftrightarrow x=\frac{-t+3}{-t-1}=\frac{t-3}{t+1}$

So $\frac{x-3}{x+1}=\frac{\frac{t-3}{t+1}-3}{\frac{t-3}{t+1}+1}=\frac{\frac{t-3-3t-3}{t+1}}{\frac{t-3+t+1}{t+1}}=\frac{-6-2t}{2t-2}=\frac{-2\left( 3+t \right)}{-2\left( -t+1 \right)}=\frac{3+t}{1-t}$

So $f\left( \frac{3+t}{1-t} \right)+f\left( t \right)=\frac{t-3}{t+1}$          (*)

Summing to get $f\left( \frac{u-3}{u+1} \right)+f\left( \frac{u+3}{1-u} \right)+2f\left( u \right)=\frac{3+u}{1-u}+\frac{u-3}{u+1}$

So $u+2f\left( u \right)=\frac{\left( 3+u \right)\left( u+1 \right)+\left( u-3 \right)\left( 1-u \right)}{1-{{u}^{2}}}$ $\Rightarrow f\left( u \right)=\frac{2{{u}^{3}}+6u}{2\left( 1-{{u}^{2}} \right)}=\frac{2u\left( {{u}^{2}}+3 \right)}{2\left( 1-{{u}^{2}} \right)}=\frac{u\left( {{u}^{2}}+3 \right)}{1-{{u}^{2}}}$