Integral Calculus Question 443
Question: $ \int_{{}}^{{}}{\frac{e^{x}\ dx}{\sqrt{1-e^{2x}}}=} $
Options:
A) $ {{\cos }^{-1}}(e^{x})+c $
B) $ -{{\cos }^{-1}}(e^{x})+c $
C) $ {{\cos }^{-1}}(e^{2x})+c $
D) $ \sqrt{1-e^{2x}}+c $
Show Answer
Answer:
Correct Answer: B
Solution:
Put $ e^{x}=t\Rightarrow e^{x}dx=dt, $ then $ \int_{{}}^{{}}{\frac{e^{x}dx}{\sqrt{1-e^{2x}}},=\int_{{}}^{{}}{\frac{dt}{\sqrt{1-t^{2}}}=-{{\cos }^{-1}}t+c}}=-{{\cos }^{-1}}(e^{x})+c $ .
BETA
