Integral Calculus Question 398
Question: $ \int_{{}}^{{}}{\frac{e^{2x}-1}{e^{2x}+1}}\ dx= $
[MP PET 1987]
Options:
A) $ \frac{e^{2x}-1}{e^{2x}+1}+c $
B) $ \log (e^{2x}+1)-x+c $
C) $ \log (e^{2x}+1)+c $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int_{{}}^{{}}{\frac{e^{2x}-1}{e^{2x}+1},dx}=\int_{{}}^{{}}{\frac{e^{x}-{e^{-x}}}{e^{x}+{e^{-x}}},dx} $ Now put $ e^{x}+{e^{-x}}=t\Rightarrow (e^{x}-{e^{-x}})dx=dt, $ then it reduces to $ \int_{{}}^{{}}{\frac{dt}{t}}=\log t=\log (e^{x}+{e^{-x}})=\log (e^{2x}+1)-x+c $ .
BETA
