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JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Critical Thinking

  • question_answer
    \[\int_{{}}^{{}}{\frac{x}{{{x}^{4}}+{{x}^{2}}+1}dx}\] equal to                [MP PET 2004]

    A) \[\frac{1}{3}{{\tan }^{-1}}\left( \frac{2{{x}^{2}}+1}{3} \right)\]

    B) \[\frac{1}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{2{{x}^{2}}+1}{\sqrt{3}} \right)\]

    C) \[\frac{1}{\sqrt{3}}{{\tan }^{-1}}(2{{x}^{2}}+1)\]

    D) None of these

    Correct Answer: B

    Solution :

    • \[I=\int_{{}}^{{}}{\frac{x}{{{x}^{4}}+{{x}^{2}}+1}\,dx=\int_{{}}^{{}}{\frac{xdx}{({{x}^{2}}+x+1)\,({{x}^{2}}-x+1)}}}\]                   
    • \[I=\frac{1}{2}\int_{{}}^{{}}{\frac{dx}{{{x}^{2}}-x+1}-\frac{1}{2}\int_{{}}^{{}}{\frac{dx}{{{x}^{2}}+x+1}}}\]                   
    • \[I=\frac{1}{2}\int_{{}}^{{}}{\frac{dx}{{{\left( x-\frac{1}{2} \right)}^{2}}+{{\left( \frac{\sqrt{3}}{2} \right)}^{2}}}-\frac{1}{2}\int_{{}}^{{}}{\frac{dx}{{{\left( x+\frac{1}{2} \right)}^{2}}+{{\left( \frac{\sqrt{3}}{2} \right)}^{2}}}}}\]                   
    • \[I=\frac{1}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{x-\frac{1}{2}}{\frac{\sqrt{3}}{2}} \right)-\frac{1}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}} \right)\]                
    • \[I=\frac{1}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{2{{x}^{2}}+1}{\sqrt{3}} \right)\].


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