JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Mock Test - Integrals

  • question_answer
    \[\int\limits_{0}^{\infty }{\left( \frac{\pi }{1+{{\pi }^{2}}{{x}^{2}}}-\frac{1}{1+{{x}^{2}}} \right)\log x\,dx}\]is equal to

    A) \[-\frac{\pi }{2}\ln \pi \]

    B) 0

    C) \[\frac{\pi }{2}\ln 2\]     

    D) none of these

    Correct Answer: A

    Solution :

    [a] \[\int\limits_{0}^{\infty }{\left( \frac{\pi }{1+{{\pi }^{2}}{{x}^{2}}}-\frac{1}{1+{{x}^{2}}} \right)\log xdx}\] \[=\int\limits_{0}^{\infty }{\frac{\log \left( \frac{y}{\pi } \right)dy}{1+{{y}^{2}}}=\int\limits_{0}^{\infty }{\frac{\log x}{1+{{x}^{2}}}dx}}\] \[=-\int\limits_{0}^{\infty }{\frac{\log \pi }{1+{{y}^{2}}}dy=-\frac{\pi }{2}In\pi }\]


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