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Thread: Improper Integral Help (Again)

  1. May 18th 2011, 05:05 AM #1
    Warrenx
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    Improper Integral Help (Again)

    Doing this problem. I have the antiderivative after using partial fraction decomposition. What I do not understand is the answer. I thought that this was divergent, but it is convergent. When I look at the end product, I do not get how that converges.

    Problem:
    50. \int^2_\infty \frac{dx}{(x+3)(x+1)^2}

    Answer:


    Now my question is, as lim R-> \infty-, does that cancel out the "ln" terms?
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  2. May 18th 2011, 05:24 AM #2
    amul28
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    are you sure the integral is \infty to 2
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  3. May 18th 2011, 05:41 AM #3
    alexmahone
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    log(x+3)-log(x+1)=log\frac{x+3}{x+1}=0 as x tends to infinity
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  4. May 18th 2011, 01:17 PM #4
    Warrenx
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    Amul you are right I put the domain backwards. Thank you Alex for that observation, totally forgot about that.
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  5. May 18th 2011, 01:57 PM #5
    Plato
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    \frac{1}{(x+3)(x+1)^2}=\frac{1}{4(x+3)}+\frac{1}{2 (x+1)^2}-\frac{1}{4(x+1)}
    Last edited by Plato; May 19th 2011 at 03:33 AM.
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  6. May 19th 2011, 02:33 AM #6
    Warrenx
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    Quote Originally Posted by Plato View Post
    \frac{1}{(x+3)(x+1)^2}=\frac{1}{4(x+1)}+\frac{1}{2 (x+1)^2}-\frac{1}{4(x+1)}
    Thank you for your help Plato .
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  7. May 19th 2011, 03:13 AM #7
    HallsofIvy
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    Quote Originally Posted by Plato View Post
    \frac{1}{(x+3)(x+1)^2}=\frac{1}{4(x+1)}+\frac{1}{2 (x+1)^2}-\frac{1}{4(x+1)}
    Am I missing something? When x= 0, the left side is \frac{1}{3} and the right side is \frac{1}{2}. I don't believe those are equal!
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  8. May 19th 2011, 03:34 AM #8
    Plato
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    Thanks. There is a typo.
    Corrected.
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