Question

∫\(\frac {e^x}{(2+e^x)(e^x +1)}\)dx = (where C is a constant of integration.)

• A. log \((\frac {e^x + 2}{e^x +1})\) + C
• B. log \((\frac {e^x}{e^x +2})\) + C
• C. \((\frac {e^x + 1}{e^x +2})\) + C
• D. log \((\frac {e^x + 1}{e^x +2})\) + C

Answer 1

The Correct Option is D

Let e^x = t Then e^x dx = dt
I = ∫\(\frac {e^x}{(2+e^x)(e^x +1)}\)dx 
I = ∫\(\frac {1}{(2+t)(t +1)}\) dt
I = ∫\((\frac {1}{t +1} - \frac {1}{2+t})\) dt
I = ∫\(\frac {1}{t +1}\) dt - ∫ \(\frac {1}{2+t}\)dt
I = log (1+t) -  log (2+t) +C
I = log \((\frac {1+t}{2+t})\) + C
Put t = e^x
I =  log \((\frac {1+e^x}{2+e^x})\) + C
I = log \((\frac {e^x+1}{e^x +2})\) + C
Therefore, the correct option is (D) log \((\frac {e^x+1}{e^x =2})\) + C