A) \[{{\cos }^{-1}}x+C\]
B) \[{{\sec }^{-1}}x+C\]
C) \[co{{t}^{-1}}x+C\]
D) \[{{\tan }^{-1}}x+C\]
Correct Answer: B
Solution :
\[\int_{{}}^{{}}{\frac{1}{x\sqrt{{{x}^{2}}-1}}dx={{\sec }^{-1}}}x+C\]You need to login to perform this action.
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