Integral Calculus Question 474
Question: $ \int_{{}}^{{}}{\frac{dx}{1-x^{2}}=} $
[MP PET 1987, 92, 2000]
Options:
A) $ {{\tan }^{-1}}x+c $
B) $ {{\sin }^{-1}}x+c $
C) $ \frac{1}{2}\ln | \frac{1+x}{1-x} |+c $
D) $ \frac{1}{2}\ln | \frac{1-x}{1+x} |+c $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \int_{{}}^{{}}{\frac{dx}{1-x^{2}}}=\frac{1}{2}\log ( \frac{1+x}{1-x} )+c=\frac{1}{2}In| \frac{1+x}{1-x} |+c. $
BETA
