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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 26

  • question_answer
    If m is a non-zero real numer such that \[\int{\frac{{{x}^{5m-1}}+2{{x}^{4m-1}}}{{{({{x}^{2m}}+{{x}^{m}}+1)}^{3}}}}dx=f(x)+C,\]where C is Constant of integration, then \[f(x)\] is

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

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

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

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

    Correct Answer: C

    Solution :

    [c] \[I=\int{\frac{{{x}^{5m-1}}+2{{x}^{4m-1}}}{{{x}^{6m}}{{(1+{{x}^{-m}}+{{x}^{-2m}})}^{3}}}}dx\] \[=\int{\frac{{{x}^{-(m+1)}}+2{{x}^{-(2m+1)}}}{{{(1+{{x}^{-m}}+{{x}^{-2m}})}^{3}}}dx}\] Putting \[(1+{{x}^{-m}}+{{x}^{-2m}})=t,\]we get \[I=\frac{-1}{m}\int{\frac{dt}{{{t}^{3}}}}=\frac{1}{2m{{t}^{2}}}+C=\frac{{{x}^{4m}}}{2m{{({{x}^{2m}}+{{x}^{m}}+1)}^{2}}}+C\]


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