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

  • question_answer
    \[\int_{{}}^{{}}{\frac{{{e}^{2x}}+1}{{{e}^{2x}}-1}\ dx}\] equals    [RPET 1996]

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

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

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

    D)            \[\log (1-{{e}^{-x}})+c\]

    Correct Answer: A

    Solution :

                       \[I=\int_{{}}^{{}}{\frac{{{e}^{2x}}+1}{{{e}^{2x}}-1}}=\int_{{}}^{{}}{\frac{{{e}^{x}}+{{e}^{-x}}}{{{e}^{x}}-{{e}^{-x}}}}\,dx\]            Put \[{{e}^{x}}-{{e}^{-x}}=t\Rightarrow ({{e}^{x}}+{{e}^{-x}})\,dx=dt\]            \[\therefore \,\,\,I=\int_{{}}^{{}}{\frac{dt}{t}\,dt}=\log t+c=\log ({{e}^{x}}-{{e}^{-x}})+c\].


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