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

Answer: (D)

log $(\frac {e^x + 1}{e^x +2})$ + C

Answer 2

Let ex = t Then ex 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 = ex
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