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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2007

  • question_answer
    The value of\[\int{{{e}^{{{\tan }^{-1}}x}}\frac{(1+x+{{x}^{2}})}{1+{{x}^{2}}}}dx\]is

    A)  \[{{\tan }^{-1}}x+c\]

    B)  \[{{e}^{{{\tan }^{-1}}x}}+2x+c\]

    C)  \[{{e}^{{{\tan }^{-1}}x}}+c\]

    D)  \[{{e}^{{{\tan }^{-1}}x}}-x+c\]

    E)  \[x{{e}^{{{\tan }^{-1}}x}}+c\]

    Correct Answer: E

    Solution :

    Let \[I=\int{{{e}^{{{\tan }^{-1}}x}}}dx+\int{{{e}^{{{\tan }^{-1}}x}}.\frac{x}{(1+{{x}^{2}})}}dx\] \[=\int{\frac{d}{dx}}(x{{e}^{{{\tan }^{-1}}x}})dx+c\] \[=x{{e}^{{{\tan }^{-1}}x}}+c\]


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