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JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Critical Thinking

  • question_answer
    \[\int_{{}}^{{}}{\frac{x-1}{{{(x+1)}^{3}}}{{e}^{x}}\ dx=}\]       [IIT 1983; MP PET 1990]

    A) \[\frac{-{{e}^{x}}}{{{(x+1)}^{2}}}+c\]

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

    C) \[\frac{{{e}^{x}}}{{{(x+1)}^{3}}}+c\]

    D) \[\frac{-{{e}^{x}}}{{{(x+1)}^{3}}}+c\]

    Correct Answer: B

    Solution :

    • \[\int_{{}}^{{}}{\frac{x-1}{{{(x+1)}^{3}}}{{e}^{x}}dx=\int_{{}}^{{}}{{{e}^{x}}\left( \frac{(x+1)}{{{(x+1)}^{3}}}-\frac{2}{{{(x+1)}^{3}}} \right)\,dx}}\]    
    • \[=\int_{{}}^{{}}{{{e}^{x}}\left( \frac{1}{{{(x+1)}^{2}}}-\frac{2}{{{(x+1)}^{3}}} \right)\,dx}=\frac{{{e}^{x}}}{{{(x+1)}^{2}}}+c\]. \[\left\{ \therefore \,\,\,\frac{d}{dx}\left( \frac{1}{{{(x+1)}^{2}}} \right)=-\frac{2}{{{(x+1)}^{3}}} \right\}\]


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