JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Self Evaluation Test - Integrals

  • question_answer
    \[\int{\frac{{{x}^{n-1}}}{{{x}^{2n}}+{{a}^{2}}}dx}=\]

    A) \[\frac{1}{na}{{\tan }^{-1}}\left( \frac{{{x}^{n}}}{a} \right)+C\]

    B) \[\frac{n}{a}{{\tan }^{-1}}\left( \frac{{{x}^{n}}}{a} \right)+C\]

    C) \[\frac{n}{a}{{\sin }^{-1}}\left( \frac{{{x}^{n}}}{a} \right)+C\]

    D) \[\frac{n}{a}{{\cos }^{-1}}\left( \frac{{{x}^{n}}}{a} \right)+C\]

    Correct Answer: A

    Solution :

    [a] Let \[I=\int{\frac{{{x}^{n-1}}dx}{{{x}^{2n}}+{{a}^{2}}}}\] Let \[{{x}^{n}}=t\Rightarrow n\,\,.\,\,{{x}^{n-1}}dx=dt\] \[\therefore I=\int{\frac{1}{n}.\frac{dt}{{{t}^{2}}+{{a}^{2}}}=\frac{1}{n}\,\,.\,\,\frac{1}{a}{{\tan }^{-1}}\left( \frac{t}{a} \right)+C}\] \[=\frac{1}{na}{{\tan }^{-1}}\left[ \frac{{{x}^{n}}}{a} \right]+C\]


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