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JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Integration by Substitution

  • question_answer
    \[\int_{{}}^{{}}{\frac{dx}{{{e}^{x}}-1}=}\]                         [MP PET 1989]

    A)            \[\ln (1-{{e}^{-x}})+c\]

    B)            \[-\ln (1-{{e}^{-x}})+c\]

    C)            \[\ln ({{e}^{x}}-1)+c\]

    D)            None of these

    Correct Answer: A

    Solution :

                       \[\int_{{}}^{{}}{\frac{dx}{{{e}^{x}}-1}=\int_{{}}^{{}}{\frac{{{e}^{-x}}}{1-{{e}^{-x}}}\,dx}}\]            Put \[1-{{e}^{-x}}=t\Rightarrow {{e}^{-x}}dx=dt,\] then it reduces to            \[\int_{{}}^{{}}{\frac{dt}{t}=\log t+c}=\log (1-{{e}^{-x}})+c.\]


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