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Math Help - Indefinite Integrals

  1. #1
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    Indefinite Integrals

    Can someone show me the steps to finding the integral of (x+1)/(x^2+2x) [ln(x^2+2x)] dx

    I'm not sure where to begin because I don't have any integration rules with the lnx
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  2. #2
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    Quote Originally Posted by john-1 View Post
    Can someone show me the steps to finding the integral of (x+1)/(x^2+2x) [ln(x^2+2x)] dx
    Do you mean 'show that' \int \frac{x+1}{x^2+2x}~dx = \ln(x^2+2x) or 'find' \int \frac{x+1}{x^2+2x}  \ln(x^2+2x)~dx

    ??
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  3. #3
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    Quote Originally Posted by pickslides View Post
    Do you mean 'show that' \int \frac{x+1}{x^2+2x}~dx = \ln(x^2+2x) or 'find' \int \frac{x+1}{x^2+2x}  \ln(x^2+2x)~dx

    ??
    the second one please
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    The second one, the first on is not so true...

    in second integral use integration by parts...
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  5. #5
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    could you show me the steps please? I only have the answer with me in my text but no steps
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  6. #6
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    Using the substitution u = \ln(x^2+2x), we have:

    \displaystyle \int\frac{(x+1)\ln(x^2+2x)}{x^2+2x}\;{dx} = \frac{1}{2}\int\left(\frac{x+1}{x^2+2x}\right)\lef  t(\frac{x^2+2x}{x+1}\right)\cdot u \;{du} = \frac{1}{2}\int{u}\;{du} = \frac{1}{4}u^2+k.

    Therefore \displaystyle \int\frac{(x+1)\ln(x^2+2x)}{x^2+2x}\;{dx} = \frac{1}{4}\ln^2(x^2+2x)+k.
    Last edited by TheCoffeeMachine; August 5th 2010 at 04:42 AM.
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  7. #7
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    TheCoffeeMachine, in your second step, why did you put (x^2 + 2x) / (x+1)? Where did that come from? mostly confused about 2nd step
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  8. #8
    Senior Member AllanCuz's Avatar
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    If  u = ln(x^2 + 2x) then  \frac{du}{dx} = \frac{2x +2}{x^2+2x} \to dx = (\frac{1}{2}) \frac{x^2 + 2x}{x+1}du

    He then subbed this into the integral for dx!
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