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

  • question_answer
    \[\int_{{}}^{{}}{{{e}^{x}}\left( \frac{1}{x}-\frac{1}{{{x}^{2}}} \right)}\,dx=\]          [AISSE 1983; MP PET 1994, 96]

    A)                 \[-\frac{{{e}^{x}}}{{{x}^{2}}}+c\]

    B)                 \[\frac{{{e}^{x}}}{{{x}^{2}}}+c\]

    C)                 \[\frac{{{e}^{x}}}{x}+c\]

    D)                 \[-\frac{{{e}^{x}}}{x}+c\]

    Correct Answer: C

    Solution :

                    \[\int_{{}}^{{}}{{{e}^{x}}\left( \frac{1}{x}-\frac{1}{{{x}^{2}}} \right)}\,dx={{e}^{x}}\frac{1}{x}+c\]                 Since, we have \[\int_{{}}^{{}}{{{e}^{x}}\left\{ f(x)+{f}'(x) \right\}\,dx}={{e}^{x}}f(x)+c.\]


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