Integral Calculus Question 451
Question: $ \int_{{}}^{{}}{\frac{e^{2x}+1}{e^{2x}-1}\ dx} $ equals
[RPET 1996]
Options:
A) $ \log (e^{x}-{e^{-x}})+c $
B) $ \log (e^{x}+{e^{-x}})+c $
C) $ \log ({e^{-x}}-e^{x})+c $
D) $ \log (1-{e^{-x}})+c $
Show Answer
Answer:
Correct Answer: A
Solution:
$ I=\int_{{}}^{{}}{\frac{e^{2x}+1}{e^{2x}-1}}=\int_{{}}^{{}}{\frac{e^{x}+{e^{-x}}}{e^{x}-{e^{-x}}}},dx $ Put $ e^{x}-{e^{-x}}=t\Rightarrow (e^{x}+{e^{-x}}),dx=dt $
$ \therefore ,I=\int_{{}}^{{}}{\frac{dt}{t},dt}=\log t+c=\log (e^{x}-{e^{-x}})+c $ .
BETA
