Integral Calculus Question 404
Question: $ \int_{{}}^{{}}{\frac{1}{{{(e^{x}+{e^{-x}})}^{2}}}\ dx=} $
Options:
A) $ -\frac{1}{2(e^{2x}+1)}+c $
B) $ \frac{1}{2(e^{2x}+1)}+c $
C) $ -\frac{1}{e^{2x}+1} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{\frac{1}{{{(e^{x}+{e^{-x}})}^{2}}},dx=\int_{{}}^{{}}{\frac{e^{2x}}{{{(e^{2x}+1)}^{2}}}},dx} $ Put $ e^{2x}+1=t\Rightarrow 2e^{2x}dx=dt, $ then it reduces to $ \frac{1}{2}\int_{{}}^{{}}{\frac{1}{t^{2}}dt}=-\frac{1}{2}.\frac{1}{t}+c=-\frac{1}{2(e^{2x}+1)}+c. $
BETA
