CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2002

  • question_answer
    \[\int{\frac{dx}{{{x}^{2}}+4x+13}}\]is equal to:

    A)  \[\log ({{x}^{2}}+4x+130)+c\]

    B)         \[\frac{1}{3}{{\tan }^{-1}}\left( \frac{x+2}{3} \right)+c\]

    C)         \[\log (2x+4)+c\]

    D)         \[\frac{1}{{{x}^{2}}+4x+13}+c\]

    E)         \[\frac{2x+4}{{{({{x}^{2}}+4x+13)}^{2}}}+c\]

    Correct Answer: B

    Solution :

    Let \[I=\int{\frac{dx}{{{x}^{2}}+4x+13}}\] \[=\int{\frac{dx}{{{x}^{2}}+4x+4+{{3}^{2}}}}=\int{\frac{dx}{{{(x+2)}^{2}}+{{3}^{2}}}}\] \[=\frac{1}{3}{{\tan }^{-1}}\left( \frac{x+2}{3} \right)+c\]


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