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JEE Main & Advanced Sample Paper JEE Main Sample Paper-27

  • question_answer
    If m is a non-zero real number then f \[\int{\frac{{{x}^{5m-1}}+2{{x}^{4m-1}}}{{{({{x}^{2m}}+{{x}^{m}}+1)}^{2}}}dx}\], is equal to [Note: C denotes constant of integration]

    A)  \[\frac{{{x}^{5m}}-{{x}^{4m}}}{2m{{({{x}^{2m}}+{{x}^{m}}+1)}^{2}}}+C\]

    B)  \[\frac{2m({{x}^{5m}}+{{x}^{4m}})}{{{({{x}^{2m}}+{{x}^{m}}+1)}^{2}}}+C\]

    C)  \[\frac{{{x}^{4m}}}{2m{{({{x}^{2m}}+{{x}^{m}}+1)}^{2}}}+C\]

    D)  \[\frac{{{x}^{5m}}}{2m.\,{{({{x}^{2m}}+{{x}^{m}}+1)}^{2}}}+C\]

    Correct Answer: C

    Solution :

    Let \[I=\int_{{}}^{{}}{\frac{({{x}^{5m-1}}\,+2{{x}^{4m-1}})dx}{{{x}^{6m}}{{(1+{{x}^{-m}}+{{x}^{-2m}})}^{3}}}}\] \[=\int_{{}}^{{}}{\frac{{{x}^{-(m+1)\,}}+2.x{{\,}^{-(2m+1)}}}{{{(1+{{x}^{-m}}+{{x}^{-2m}})}^{3}}}dx}\] Put \[(1+{{x}^{-m}}+{{x}^{-2m}})=t\] \[I=\frac{-1}{m}\int_{{}}^{{}}{\frac{dt}{{{t}^{3}}}\,=\frac{1}{2m{{t}^{2}}}+C=\,\frac{{{x}^{4m}}}{2m{{({{x}^{2m}}+{{x}^{m}}+1)}^{2}}}+C}\]


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