Integral Calculus Question 40
Question: $ \int_{{}}^{{}}{\frac{dx}{e^{x}+1-2{e^{-x}}}=} $
Options:
A) $ \log (e^{x}-1)-\log (e^{x}+2)+c $
B) $ \frac{1}{2}\log (e^{x}-1)-\frac{1}{3}\log (e^{x}+2)+c $
C) $ \frac{1}{3}\log (e^{x}-1)-\frac{1}{3}\log (e^{x}+2)+c $
D) $ \frac{1}{3}\log (e^{x}-1)+\frac{1}{3}\log (e^{x}+2)+c $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \int_{{}}^{{}}{\frac{e^{x}dx}{e^{2x}+e^{x}-2}}=\int_{{}}^{{}}{\frac{dt}{t^{2}+t-2}} $ $ { \because ,e^{x}=t\Rightarrow e^{x}dx=dt } $ $ =\int_{{}}^{{}}{\frac{dt}{(t+2)(t-1)}}=,\int_{{}}^{{}}{\frac{1}{3}[ \frac{1}{t-1}-\frac{1}{t+2} ]},dt $ $ =\frac{1}{3}\log (e^{x}-1)-\frac{1}{3}\log (e^{x}+2)+c. $
BETA
