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Thread: Integration

  1. #1
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    Integration

    how do you integrate x-1/x+1 dx from 0 to 1


    PLeaSe HeLp!!!!!!!!!!!!!!!
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  2. #2
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    Quote Originally Posted by sheven
    how do you integrate x-1/x+1 dx from 0 to 1

    PLeaSe HeLp!!!!!!!!!!!!!!!
    Hello,

    I assume that you want to calculate $\displaystyle \int^{1}_{0} \frac{x-1}{x+1} dx$ from 0 to 1.

    First do a long division: $\displaystyle \frac{x-1}{x+1} =1-\frac{2}{x+1}$.

    And now you get: $\displaystyle \int^{1}_{0} \frac{x-1}{x+1} dx=\left[x-2\cdot \ln(x+1) \right]^{1}_{0}$

    Put in the borders:

    $\displaystyle \left( 1-2\cdot \ln(2) \right)-\left(0-2\cdot \ln(1) \right)=1-2\cdot \ln(2)$

    Greetings

    EB
    Last edited by earboth; Feb 22nd 2006 at 08:47 PM.
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  3. #3
    TD!
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    Instead of doing a long division, remember this trick:

    $\displaystyle \frac{{x - 1}}{{x + 1}} = \frac{{x + 1 - 2}}{{x + 1}} = \frac{{x + 1}}{{x + 1}} - \frac{2}{{x + 1}} = 1 - \frac{2}{{x + 1}}$
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  4. #4
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    I have one more method
    Instead of doing long divison or the trick. You can use substitution,
    $\displaystyle u=x+1$, then $\displaystyle u-2=x-1$
    Thus, by the substitution rule, (note that $\displaystyle \frac{du}{dx}=1$
    $\displaystyle \int\frac{u-2}{u}{du}$
    Thus,
    $\displaystyle \int u-\frac{2}{u}du$
    Thus,
    $\displaystyle \frac{1}{2}u^2-2\ln |u| +C$
    Substitute back,
    $\displaystyle \frac{1}{2}(x+1)^2-2\ln |x+1|+C$.
    Q.E.D.
    Last edited by ThePerfectHacker; Feb 23rd 2006 at 12:21 PM.
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