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

  • question_answer
    \[\int{{{\left( x+\frac{1}{x} \right)}^{n+5}}\left( \frac{{{x}^{2}}-1}{{{x}^{2}}} \right)dx}\] is equal to:

    A) \[\frac{{{\left( x+\frac{1}{x} \right)}^{n+6}}}{n+6}+c\]

    B) \[{{\left[ \frac{{{x}^{2}}+1}{{{x}^{2}}} \right]}^{n+6}}(n+6)+c\]

    C) \[{{\left[ \frac{x}{{{x}^{2}}+1} \right]}^{n+6}}(n+6)+c\]

    D) None of these

    Correct Answer: A

    Solution :

    [a] \[I=\int{{{\left( x+\frac{1}{x} \right)}^{n+5}}\left( \frac{{{x}^{2}}-1}{{{x}^{2}}} \right)dx}\] Put \[x+\frac{1}{x}=t\Rightarrow \left( 1-\frac{1}{{{x}^{2}}} \right)dx=dt\] \[\Rightarrow \left( \frac{{{x}^{2}}-1}{{{x}^{2}}} \right)dx=dt\] \[\therefore I=\int{{{t}^{n+5}}dt=\frac{{{t}^{n+6}}}{n+6}+c=\frac{{{\left( x+\frac{1}{x} \right)}^{n+6}}}{n+6}+c}\]


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