I read this as ln(1+x) where ln means log to base e (natural logarithm).
Let f⁽⁰⁾(x)=f(x)=ln(1+x); f⁽⁰⁾(0)=ln(1)=0
f⁽¹⁾(x)=1/(1+x); f⁽¹⁾(0)=1
f⁽²⁾(x)=-1/(1+x)²; f⁽²⁾(0)=-1
f⁽³⁾(x)=2!/(1+x)³; f⁽³⁾(0)=2!
f⁽⁴⁾(x)=-3!/(1+x)⁴; f⁽⁴⁾(0)=-3!
...
f⁽ⁿ⁾(x)=(-1)ⁿ⁻¹(n-1)!/(1+x)ⁿ; f⁽ⁿ⁾(0)=(-1)ⁿ⁻¹(n-1)!
The general Maclaurin term is (-1)ⁿ⁻¹(n-1)!xⁿ/n!=(-1)ⁿ⁻¹xⁿ/n.
The series is x-x²/2+x³/3-x⁴/4+...