The function f(t)=log(1+t) is defined over [0,x] and differentiable over ]0,x[, so by the mean value theorem there exists c in ]0,x[ such that f(x)-f(0)=(x-0)f'(c), that is log(1+x)=x/(1+c).
0<c<x implies 1<1+c<1+x and so1/(1+x)<1/(1+c)<1, but x>0 it follows that x/(1+x)<x/(1+c)<x and so
x/(1+x)<log(1+x)<x.