CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2007

  • question_answer
    The value of\[\int{\frac{{{e}^{5{{\log }_{e}}x}}-{{e}^{4{{\log }_{e}}x}}}{{{e}^{3{{\log }_{e}}x}}-{{e}^{2{{\log }_{e}}x}}}}dx\]is

    A)  \[{{x}^{2}}+c\]                 

    B)  \[\frac{{{x}^{2}}}{2}+c\]

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

    D)         \[\frac{x}{2}+c\]

    E)  \[e\]

    Correct Answer: C

    Solution :

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


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