If $f(x)= \ln\left ( \frac{1+x}{1-x} \right )$ then
$A.\,f(x_1)\cdot f(x_2)= f(x_1+x_2)\\ B.\,f(x+2)-2f(x+1)+f(x)=0\\ C.\,f(x)+f(x+1)=f(x^2+x)\\ D.\,f(x_1)+f(x_2)=f\left(\frac{x_1+x_2}{1+x_1 x_2}\right) $
As you can see, there four options and in the exam we get only 2–3 minutes to solve and I can't keep checking each and every option. Is there any other option?
Given your constraints, $-1<x<1$
Set $x_1=x_2=0$ and try them.
Then: $$f(x)=\ln(\frac 11)=0$$ $f(x+1)$ is undefined.
$f(x+2)$ is also undefined.
This eliminates $B$ and $C$.
Now to check $A$ and $D$.
$A$ works, and so does $D$
Now let $x_1=\frac 12, x_2=0$
We have that $f(x_1)=\ln 3, f(x_2)=0, f(x_1+x_2)=\ln3$, and so $A$ is false, leaving only $D$