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Math Help - Solving Improper Integrals of type 1

  1. #1
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    Solving Improper Integrals of type 1

    I ve been trying to solve this integral where it reads

    int (-infinity to +infinity) x/((x^2+4)^3/2) so i wrote this as a limit where
    lim A approaches -infinity (int from A to 0 x/((x^2+4)^3/2) + lim B approaches +infinity (int from 0 to B x/((x^2+4)^3/2). By subbing in u=x^2+4 i got the integral to equal -1/sqrt (X^2+4)+c and ended up with an answer of 0 when both integrals were added. is this correct i feel like this is wrong somehow
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  2. #2
    Senior Member yeKciM's Avatar
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    you got correct result it's zero

     \displaystyle \int \frac {x\; dx }{\sqrt{(x^2+4)^{3}} } = -\frac {1}{\sqrt{(x^2+4)}} \big{|} _{-\infty} ^{+\infty} = 0
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  3. #3
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    That's correct.
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