If f (x) = 1+x/1-x then f (2x/1+ x^2) is equal to --------------

f(x)=log(1+x)/(1-x)

Replace x with (2*x)/(1+x^2)

f(2*x/1+x^2)=log(1+(2*x)/(1+x^2))-log(1-(2*x)/(1+x^2))

=> {log(1+x^2+2*x)-log(1+x^2)}-{log(1+x^2–2*x)-log(1+x^2)}

=>log(1+x^2+2*x)-log(1+x^2–2*x)

=>log(x+1)^2-log(x-1)^2

=>2{log(x+1)-log(x-1)}

=>2{log((x+1)/(x-1))}

=>2{f(x)}

Hence

f(2*x/1+x^2)=2*f(x)