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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2001

  • question_answer
    \[\int_{0}^{1}{\frac{{{\tan }^{-1}}x}{1+{{x}^{2}}}dx}\] is equal to:

    A) \[\pi /4\]                             

    B)  \[{{\pi }^{2}}/32\]

    C)  1                                            

    D)  none of these

    Correct Answer: B

    Solution :

    \[\int_{0}^{\pi }{\frac{{{\tan }^{-1}}x}{1+{{x}^{2}}}dx}\]put \[{{\tan }^{-1}}x=t\] \[\frac{1}{1+{{x}^{2}}}dx=dt\] \[\int_{0}^{\pi /4}{t\,dt=\left[ \frac{{{t}^{2}}}{2} \right]}_{0}^{\pi /4}={{\pi }^{2}}/32\]


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