This can be solved using partial fractions and substitution.
Substitute ex=t.exdx=dt
Integral becomes ∫1(1+t)(2+t)dt
Using partial fractions,
1(1+t)(2+t)=2(1+t)−(2+t)(1+t)(2+t)=22+t−11+t
∫22+tdt=2ln|t+2|
∫11+tdt=ln|t+1|
∫1(1+t)(2+t)dt=2ln|t+2|−ln|t+1|+c