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How do I go about integrating this?

  1. Jan 13, 2010 #1
    1. The problem statement, all variables and given/known data

    Find [tex]\int[/tex][tex]\frac{x+2}{x^{2}+2*x+10}[/tex]*dx

    You are given that [tex]\int[/tex][tex]\frac{du}{u^{2}+a^{2}}[/tex] = [tex]\frac{1}{a}[/tex]tan[tex]^{-1}[/tex] [tex]\frac{u}{a}[/tex] + C for a not equal to 0

    2. Relevant equations

    None.

    3. The attempt at a solution

    I'm not entirely too sure how to go about doing this. I first thought of integration by parts, but given that this is a 3 mark question it seems quite long winded.

    I split up the expression:

    1/x^2+2*x+10*x+2*dx

    Then I set u = x+2, du = 1, dv =1/x^2+2*x+10, v = (1/3)*arctan((1/3)*x+1/3)

    I have no idea how to get v, that was calculated using Maple, so how is that arctan component calculated by hand?

    Using the uv - int(vdu) formula for the rest of it results in some bizarre answers, which might be right but seems altogether too long and complex for a 3 mark answer.


    The final answer as calculated by Maple is:

    (1/2)*ln(x^2+2*x+10)+(1/3)*arctan((1/3)*x+1/3)

    Any help and direction greatly appreciated.
     
  2. jcsd
  3. Jan 13, 2010 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Please use parentheses to show what you mean! I finally realized that this was (1/(x^2+ 2x+ 10))(x+2). What you wrote would be more readily interpreted as 1/(x^2+ 2x+ 10x+ 2).

    Don't do this "by parts", complete the square in the denominator. [itex]x^2+ 2x+ 10= x^2+ 2x+ 1- 1+10= (x+1)^2+ 9[/itex]. Now you can let u= x+1 so that du= dx and x= u- 1. The integral becomes
    [tex]\int \frac{u- 1}{u^2+ 9} du= \int \frac{u}{u^2+9} du- \int \frac{1}{u^2+9}du[/tex]
    The first integral can be done with the substitution [itex]v= u^2+9[/itex] and 1/(u^2+9) is exactly what you need for your arctan integral.

     
  4. Jan 13, 2010 #3
    HallsofIvy, thank you so much for your response, that was a lot of help!

    Sorry about the lack of parentheses.
     
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