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I'm really having a problem integrating this equation, infact i have no idea

  1. Feb 25, 2009 #1
    1. The problem statement, all variables and given/known data

    Please tell me how to integrate this...

    2. Relevant equations

    [(2x + 4)][(2x^2 + 3x + 1)^(1/2)]

    3. The attempt at a solution
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Feb 25, 2009 #2


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    [itex]2x^2+ 3x+ 1= (2x+1)(x+1)
    Complete the square in the square root: [itex]2x^2+ 3x+ 1= 2(x^2+ (3/2)x+ 9/16)+1-9/8= 2(x+ 3/2)^2- 1/8[/itex]. Let u= x+ 3/2, then du= dx and x= u- 3/2 so that 2x+ 4= 2u+ 1 so the function to be integrated becomes [itex](2u+1)(2u^2+ 7/16)^{1/2}= 2\sqrt{2}u(u^2+ 7/32)^{1/2}+ \sqrt{2}(u^2+ 7/32}^{1/2}[/itex]. The first can be integrated by the substitution [itex]v= u^2+ 7/32[/itex] and the second by a trig substitution.
    Last edited: Feb 27, 2009
  4. Feb 27, 2009 #3
    I get the idea but you have made one mistake; it should be 9/8 instead of 9/16 outside the bracket. Thanks alot though
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