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Help with integration problem

  1. Feb 23, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex]\int\frac{x+1}{x^2+2x}[/tex]


    2. Relevant equations

    N/A

    3. The attempt at a solution

    I'm not sure what method to use.


    Thank you.
     
  2. jcsd
  3. Feb 23, 2010 #2
    Re: Integral

    You have to use partial fraction integration which involves splitting the integral into terms that are easier to integrate. After pulling an "x" from the bottom of the original integral you get INT (x+1)/x(x+2)

    After splitting and such:

    x + 1 = A(x+2) + Bx

    Plugging in clever values for x we will get that A & B both equal 1/2

    Substituting back into our original problem we will have the INT of .5/x + .5/(x+2) which will yield an answer of 0.5 ln|x| + 0.5 ln|x+2|

    I'm assuming you know how to use partial fraction integration. If not, respond back and i'll try helping you further.

    Hope that helps!
     
  4. Feb 23, 2010 #3

    Mentallic

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    Re: Integral

    Try taking the derivative of the denominator :wink:
     
  5. Feb 23, 2010 #4

    Mentallic

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    Re: Integral

    IntegrateMe's answer is the same as what you would get if following my advice. By the way, that method is unnecessarily complicated :tongue:
     
  6. Feb 23, 2010 #5
    Re: Integral

    Haha, Mentallic, i don't really know any other way to do it.

    Can you please explain a simpler way, i'd like to know how to solve these problems quicker!
     
  7. Feb 23, 2010 #6

    Mentallic

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    Re: Integral

    You've been taught partial fractions but not u-substitution?

    In any case, let u=x2+2x

    Then du/dx=2x+2 and du=2(x+1)dx

    Substitute u and du back into the integral accordingly - remember that there is no 2 in the integral so you'll need to make it du/2=(x+1)dx. Now it's simple to integrate (and don't forget to substitute u=x2+2x back into your answer!)
     
  8. Feb 23, 2010 #7
    Re: Integral

    ^_^

    It's strange having an "x" in your "u-value" but i guess that way is easier. Thanks :)
     
  9. Feb 23, 2010 #8
    Re: Integral

    Thank you very much!
     
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