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Integrate: dx/sqrt(x^2+2x +5)

  1. Mar 17, 2007 #1
    1. The problem statement, all variables and given/known data

    Integrate: dx/sqrt(x^2+2x +5)



    2. Relevant equations
    refer to above

    3. The attempt at a solution

    I can integrate the equation dx/sqrt(1+x^2) using the rules

    cosh^2 u - sinh^2 u = 1
    cosh^2 u = 1+sinh^2 u

    but i dont know where to start with this question because of the 2x + 5 in the denomentor. Could someone please point me in the right direction.
     
  2. jcsd
  3. Mar 17, 2007 #2

    George Jones

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    Complete the square.
     
  4. Mar 17, 2007 #3
    Thanks that helped but i'm still stuck with a 4 i don't know to get rid of :S

    integral of: dx/sqrt(x^2+2x+5)

    Equals the integral of: dx/sqrt((x+1)^2 + 4)

    Using:
    1. (x+1) = sinh u
    2. cosh^2 u = 1 + sinh^2 u
    3. dx = cosh u du

    I get to this stage:

    Integral of: dx/sqrt((x+1)^2 + 4)

    Equals the integral of: cosh u du/((x+1)^2 + 4)

    This is where i get stuck, i'm not sure what to do with the 4. could som1 plz help.

    thanx again George
     
  5. Mar 17, 2007 #4

    George Jones

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    In an appropriate manner, take the 4 outside the square root.
     
  6. Mar 17, 2007 #5
    ya i squared everything but the bottom line is not in the correct form.

    i hav:

    cosh ^2 u du / (x+1)^2 +4

    if the denomentator was (x+1)^2 + 1 i could just substitute but cos of that 4 i can't. gettin rid of taht 4 is my problem.
     
  7. Mar 17, 2007 #6

    George Jones

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    Factor the 4 out of both terms that are inside the square root.

    Hint: 1 = 4 * 1/4.
     
  8. Mar 17, 2007 #7

    matt grime

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    There is more than one possible substitution you can make, you know.
     
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