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Completing the square and then using trig substitutio

  1. Mar 2, 2009 #1
    1. The problem statement, all variables and given/known data
    Integral of 1/[(x^2+4x+3)^(3/2)]

    3. The attempt at a solution
    I tried completing the square and then using trig substitution. So its:

    1/[(x+2)^2 -1] let x+2=sec(theta), dx =sec(theta)tan(theta)

    After some simplifying, I get it down to integral of csc(theta) which = -ln|csc(theta) + cot(theta)|

    My calc teacher has the answers online, and it doesnt have any ln in it at all, so im pretty sure im on the wrong track here. Any suggestions?
     
  2. jcsd
  3. Mar 2, 2009 #2

    Gib Z

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    Homework Helper

    Re: integral

    It seems like you completely forgot about the exponent in the denominator.
     
  4. Mar 2, 2009 #3
    Re: integral

    The small mistake you made after completing the square is that you forgit to copy down the ^3/2 in the denominator. It should be:

    [tex]\frac{1}{[(x+2)^{2} -1]^{\frac{3}{2}}}[/tex].

    It becomes much easier when you include it ;). But aside from that mistake you're on teh right general track.
     
    Last edited: Mar 2, 2009
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