Integral Calculus Question 39
Question: $ \int_{{}}^{{}}{\frac{e^{x}}{(1+e^{x})(2+e^{x})}dx=} $
Options:
A) $ \log [(1+e^{x})(2+e^{x})]+c $
B) $ \log [ \frac{1+e^{x}}{2+e^{x}} ]+c $
C) $ \log [(1+e^{x})\sqrt{2+e^{x}}]+c $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int_{{}}^{{}}{\frac{e^{x}}{(1+e^{x})(2+e^{x})},dx}=\int_{{}}^{{}}{{ \frac{e^{x}}{1+e^{x}}-\frac{e^{x}}{2+e^{x}} }dx} $ Now put $ 1+e^{x}=t $ and $ 2+e^{x}=t, $ then the required integral $ =\log (1+e^{x})-\log (2+e^{x})=\log ( \frac{1+e^{x}}{2+e^{x}} )+c. $
BETA
