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Math Help - Natural Logarithm Integration

  1. #1
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    Natural Logarithm Integration

    Integrate (1/x+2), where x= -6, x= -3

    Please Help
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  2. #2
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    Quote Originally Posted by creatively12 View Post
    Integrate (1/x+2), where x= -6, x= -3

    Please Help
    I assume you're asking to find

    \int_{-6}^{-3}{\frac{1}{x + 2}\,dx}

     = \left[\ln{|x + 2|}\right]_{-6}^{-3}

     = \ln{|-6 + 2|} - \ln{|-3 + 2|}

     = \ln{|-4|} - \ln{|-1|}

     = \ln{4} - \ln{1}

     = \ln{(2^2)} - 0

     = 2\ln{2}.
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  3. #3
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    Thanks man, but the answer is -1n(4)
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  4. #4
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    The 3 line is
    <br />
\ln | -3+2 | - \ln | -6 +2 |<br />
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  5. #5
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    Let u = x+2, then \dfrac{du}{dx} = 1 \Rightarrow {dx} = {du}. When x = -3, u = -1; when x = -6, u = -4. So \int_{-6}^{-3}\dfrac{1}{x+2}\;{dx} = \int_{-4}^{-1}\dfrac{1}{u}\;{du} =  \bigg[\ln{|u|}\bigg]_{-4}^{-1} = \ln\left(1\right)-\ln\left(4\right) = \boxed{-\ln\left(4\right)}  , since \ln\left(1\right) = 0.
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