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JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Critical Thinking

  • question_answer
    If \[\int_{{}}^{{}}{\frac{2x+3}{(x-1)({{x}^{2}}+1)}\ dx={{\log }_{e}}\left\{ {{(x-1)}^{\frac{5}{2}}}{{({{x}^{2}}+1)}^{a}} \right\}}-\frac{1}{2}{{\tan }^{-1}}x+A\]  , where A is any arbitrary constant, then the value of ?a? is  [MP PET 1998]

    A) 5/4

    B) - 5/3

    C) - 5/6

    D) - 5/4

    Correct Answer: D

    Solution :

    • \[I=\int_{{}}^{{}}{\frac{2x+3}{(x-1)({{x}^{2}}+1)}\,dx}\]                     
    • \[=\int_{{}}^{{}}{\frac{5\,dx}{2\,(x-1)}}+\int_{{}}^{{}}{\frac{-\left( \frac{5}{2}x+\frac{1}{2} \right)}{{{x}^{2}}+1}}\,dx\], (By partial fraction)                   
    • \[I=\frac{5}{2}\log (x-1)-\frac{5}{2}\int_{{}}^{{}}{\frac{x\,dx}{1+{{x}^{2}}}-\frac{1}{2}\int_{{}}^{{}}{\frac{dx}{1+{{x}^{2}}}}}\]                   
    • \[I=\frac{5}{2}\log (x-1)-\frac{5}{4}\log (1+{{x}^{2}})-\frac{1}{2}{{\tan }^{-1}}x+A\]                   
    • \[I=\log {{(x-1)}^{5/2}}{{(1+{{x}^{2}})}^{-5/4}}-\frac{1}{2}{{\tan }^{-1}}x+A\]                
    • On comparing, \[a=-\frac{5}{4}.\]


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