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JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Critical Thinking

  • question_answer
    If \[\int_{{}}^{{}}{\frac{4{{e}^{x}}+6{{e}^{-x}}}{9{{e}^{x}}-4{{e}^{-x}}}dx=Ax+B\log (9{{e}^{2x}}-4)}+C\], then A, B and C are           [IIT 1990]

    A) \[A=\frac{3}{2},\ B=\frac{36}{35},\ C=\frac{3}{2}\log 3+\]constant   

    B) \[A=\frac{3}{2},\ B=\frac{35}{36},\ C=\frac{3}{2}\log 3+\]constant

    C) \[A=-\frac{3}{2},\ B=-\frac{35}{36},\ C=-\frac{3}{2}\log 3+\]constant

    D) None of these

    Correct Answer: D

    Solution :

    • \[I=\int_{{}}^{{}}{\frac{4{{e}^{x}}+6{{e}^{-x}}}{9{{e}^{2x}}-4{{e}^{-x}}}\,dx}=\frac{4}{9}\int_{{}}^{{}}{\frac{9{{e}^{2x}}dx}{9{{e}^{2x}}-4}+6}\int_{{}}^{{}}{\frac{dx}{9{{e}^{2x}}-4}}\]                   
    • \[\therefore \,\,\,\int_{{}}^{{}}{\frac{dx}{9{{e}^{2x}}-4}}=\frac{1}{8}\log (9{{e}^{2x}}-4)-\frac{1}{4}\log 3-\frac{1}{4}x+\]const.                   
    • \[\therefore \,\,\,I=\frac{35}{36}\log (9{{e}^{2x}}-4)-\frac{3}{2}x-\frac{3}{2}\log 3+\]const.                   
    • Comparing with the given integral, we get                
    • \[A=-\frac{3}{2},\] \[B=\frac{35}{36},\] \[C=-\frac{3}{2}\log 3+\] const.


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