If the primitive of 1ex−12 isf(x)−log⁡|g(x)|+C then

∫1ex−12dx=∫exdxexex−12=∫dtt(t−1)2t=ex=∫1t−11t−1−1tdt=∫1(t−1)2−1t−1+1tdt=11−t−log⁡|t−1|+log⁡|t|+C=1−ex−1−log⁡1−e−x+CHence f(x)=1−ex−1 and g(x)=1−e−xThe domain of f=R~{0}