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Thread: Improper Integral Help (Again)

  1. #1
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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. $\displaystyle \int^2_\infty \frac{dx}{(x+3)(x+1)^2}$

    Answer:


    Now my question is, as lim R-> $\displaystyle \infty-$, does that cancel out the "ln" terms?
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  2. #2
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    are you sure the integral is $\displaystyle \infty $ to $\displaystyle 2$
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  3. #3
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    $\displaystyle log(x+3)-log(x+1)=log\frac{x+3}{x+1}=0$ as x tends to infinity
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  4. #4
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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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    $\displaystyle \frac{1}{(x+3)(x+1)^2}=\frac{1}{4(x+3)}+\frac{1}{2 (x+1)^2}-\frac{1}{4(x+1)}$
    Last edited by Plato; May 19th 2011 at 03:33 AM.
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  6. #6
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    Quote Originally Posted by Plato View Post
    $\displaystyle \frac{1}{(x+3)(x+1)^2}=\frac{1}{4(x+1)}+\frac{1}{2 (x+1)^2}-\frac{1}{4(x+1)}$
    Thank you for your help Plato .
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  7. #7
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    Quote Originally Posted by Plato View Post
    $\displaystyle \frac{1}{(x+3)(x+1)^2}=\frac{1}{4(x+1)}+\frac{1}{2 (x+1)^2}-\frac{1}{4(x+1)}$
    Am I missing something? When x= 0, the left side is $\displaystyle \frac{1}{3}$ and the right side is $\displaystyle \frac{1}{2}$. I don't believe those are equal!
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  8. #8
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    Thanks. There is a typo.
    Corrected.
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