Integral Calculus Question 436
Question: $ \int_{{}}^{{}}{\frac{dx}{{e^{-2x}}{{(e^{2x}+1)}^{2}}}=} $
Options:
A) $ \frac{-1}{2(e^{2x}+1)}+c $
B) $ \frac{1}{2(e^{2x}+1)}+c $
C) $ \frac{1}{e^{2x}+1}+c $
D) $ \frac{-1}{e^{2x}+1}+c $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{\frac{dx}{{e^{-2x}}{{(e^{2x}+1)}^{2}}}}=\int_{{}}^{{}}{\frac{e^{2x}dx}{{{(e^{2x}+1)}^{2}}}} $ Put $ t=e^{2x}+1\Rightarrow \frac{dt}{2}=e^{2x}dx, $ then it reduces to $ \frac{1}{2}\int_{{}}^{{}}{\frac{1}{t^{2}}},dt=-\frac{1}{2t}+c=\frac{-1}{2(e^{2x}+1)}+c. $
BETA
