A) \[{{e}^{x}}+c\]
B) \[({{e}^{x}}+1)+c\]
C) \[\log ({{e}^{x}}+1)+c\]
D) None of these
Correct Answer: C
Solution :
Put \[{{e}^{x}}+1=t\Rightarrow {{e}^{x}}dx=dt\] \[\therefore \] \[\int_{{}}^{{}}{\frac{{{e}^{x}}}{{{e}^{x}}+1}dx=\int_{{}}^{{}}{\frac{dt}{t}=\log t+c=\log ({{e}^{x}}+1)+c}}\].
You need to login to perform this action.
You will be redirected in
3 sec
