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JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Fundamental Integration

  • question_answer
    \[\int_{{}}^{{}}{\frac{2x}{{{(2x+1)}^{2}}}dx=}\] [DSSE 1985]

    A)            \[\frac{1}{2}\log (2x+1)+\frac{1}{2(2x+1)}+c\]                           

    B)            \[\frac{1}{2}\log (2x+1)-\frac{1}{2(2x+1)}+c\]

    C)            \[2\log (2x+1)+\frac{1}{2(2x+1)}+c\]                                          

    D)            \[2\log (2x+1)-\frac{1}{2(2x+1)}+c\]

    Correct Answer: A

    Solution :

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


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