Question
Mathematics Question on integral
Integrate the function: (1+ex)(2+ex)ex
Answer
(1+ex)(2+ex)ex
Let ex=t⇒exdx=dt
⇒∫(1+ex)(2+ex)exdx = ∫(t+1)(t+2)dt
=$∫[\frac {1}{(t+1)}-\frac {1}{(t+2)}]dt$
= $log\ |t+1|-log\ |t+2|+C$
= $log|\frac {t+1}{t+2}|+C$
= $log|\frac {1+e^x}{2+e^x}|+C$