A) \[{{\tan }^{-1}}x+c\]
B) \[{{\sin }^{-1}}x+c\]
C) \[\frac{1}{2}\ln \left| \frac{1+x}{1-x} \right|+c\]
D) \[\frac{1}{2}\ln \left| \frac{1-x}{1+x} \right|+c\]
Correct Answer: C
Solution :
\[\int_{{}}^{{}}{\frac{dx}{1-{{x}^{2}}}}=\frac{1}{2}\log \left( \frac{1+x}{1-x} \right)+c=\frac{1}{2}\,\,\text{In}\,\,\left| \frac{1+x}{1-x} \right|+c.\]
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