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