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

  • question_answer
    \[\int_{{}}^{{}}{\frac{{{e}^{x}}(x-1)}{{{x}^{2}}}\ dx=}\]

    A)                 \[\frac{1}{x}{{e}^{x}}+c\]

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

    C)                 \[\frac{1}{{{x}^{2}}}{{e}^{x}}+c\]

    D)                 \[\left( x-\frac{1}{x} \right){{e}^{x}}+c\]

    Correct Answer: A

    Solution :

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


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