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BCECE Engineering BCECE Engineering Solved Paper-2012

  • question_answer
    \[\int_{{}}^{{}}{\frac{1}{x\sqrt{{{x}^{2}}-1}}dx}\]is equal to

    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 :

    \[\int_{{}}^{{}}{\frac{1}{x\sqrt{{{x}^{2}}-1}}dx={{\sec }^{-1}}}x+C\]


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