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JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Critical Thinking

  • question_answer
    \[\int_{{}}^{{}}{\frac{{{x}^{2}}+1}{{{x}^{4}}-{{x}^{2}}+1}\ dx=}\]         [MP PET 1991]

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

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

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

    D) \[{{\cot }^{-1}}\left( \frac{{{x}^{2}}-1}{x} \right)+c\]

    Correct Answer: C

    Solution :

    • \[\int_{{}}^{{}}{\frac{{{x}^{2}}+1}{{{x}^{4}}-{{x}^{2}}+1}\,dx=\int_{{}}^{{}}{\frac{\left( 1+\frac{1}{{{x}^{2}}} \right)}{{{x}^{2}}+\frac{1}{{{x}^{2}}}-1}}}\]                   
    • \[=\int_{{}}^{{}}{\frac{1+\frac{1}{{{x}^{2}}}}{{{\left( x-\frac{1}{x} \right)}^{2}}+1}\,dx=\int_{{}}^{{}}{\frac{dt}{{{t}^{2}}+1}}={{\tan }^{-1}}t+c}\] \[\left\{ \text{Putting}\,x-\frac{1}{x}=t\Rightarrow \left( 1+\frac{1}{{{x}^{2}}} \right)\,dx=dt \right\}\]                
    • \[={{\tan }^{-1}}\left( x-\frac{1}{x} \right)+c={{\tan }^{-1}}\left( \frac{{{x}^{2}}-1}{x} \right)+c.\]


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