The original problem is to verify x>0 and f(x)>0;

g(x)= f(x)*( 1+x)= ln(x+ 1)*( 1+x)-(arctanx)

Because x+ 1 is greater than 0, the original problem is to prove that g(x)>0.

Derive the first derivative g' (x) = ln (x+1)+1-1(1+x? )

Second derivative g'' (x) =1(x+1)+2x/(1+x? )?

Because of x>0, the second derivative g'' (x) >: 0,

So the first derivative is increasing function, and the minimum value is obtained when x=0, that is, g'(0)=0.

Because of x>0, the first derivative g (x) >; 0.

So the original function g(x) is increasing function,

When x=0, take the minimum value g(0)=0.

G(x)>0 has been established because of x>0.

That is ln (x+1) * (1+x)-(arctanx) >; 0

Therefore, when x is greater than 0, ln(x+ 1) is greater than (arctanx)÷( 1+x).