CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    \[\int{\frac{{{e}^{6{{\log }_{e}}x}}-{{e}^{5{{\log }_{e}}x}}}{{{e}^{4{{\log }_{e}}x}}-{{e}^{3{{\log }_{e}}x}}}}dx\]is equal to

    A)  \[\frac{{{x}^{3}}}{3}+c\]                              

    B)  \[\frac{{{x}^{2}}}{2}+c\]

    C)  \[\frac{{{x}^{2}}}{3}+c\]              

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

    E)  \[x+c\]

    Correct Answer: A

    Solution :

    \[I=\int{\frac{{{e}^{6{{\log }_{e}}x}}-{{e}^{-5{{\log }_{e}}x}}}{{{e}^{4{{\log }_{e}}x}}-{{e}^{3{{\log }_{e}}x}}}}dx\] \[=\int{\frac{{{x}^{6}}-{{x}^{5}}}{{{x}^{4}}-{{x}^{3}}}}dx=\int{\frac{{{x}^{5}}(x-1)}{{{x}^{3}}(x-1)}}dx\] \[=\int{{{x}^{2}}dx}=\frac{{{x}^{3}}}{3}+c\]         


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