f(x)=ln[(x+1)/x]
f'(x)=x/(x+1)*[(x+1)/x]'
=x/(x+1)*(-1/x²)
=-1/(x²+x)
f(x)=[ln(x+1)]/x
f'(x)=[x/(x+1)-ln(x+1)]/x²
=1/(x²+1)-[ln(x+1)]/x²