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Integral I don't understand?

  1. Sep 29, 2011 #1
    1. The problem statement, all variables and given/known data
    [tex]\int \frac{2x}{3x^{2}+10x+3} dx[/tex]
    2. Relevant equations

    3. The attempt at a solution

    I can't think of a U-substitution that would work, nor a trigonometric substitution, or integration by part.

    [tex]\int \frac{2x}{3x^{2}+10x+3} dx[/tex]
    [tex]\int \frac{2x}{(x+3)(3x+1)} dx[/tex]

    I factored the denominator out thinking that I could somehow substitute for one product, but that doesn't work clearly. How do you integrate functions like these??

    I popped it into wolfram and it had a step about fractional decomposition, but I am having a hard time understanding it and we have not covered it yet in my course.

    Here is my go at it:
    It has to be in this form right?
    [tex]\frac{2x}{(3+x)(3x+1)} = \frac{A}{3+x} + \frac{B}{3x+1}[/tex]

    So now I would multiply the LCD through the equation leaving:
    [tex]2x = A(3x+1) + B(3+x)[/tex]

    I don't understand what to do now though?
     
  2. jcsd
  3. Sep 29, 2011 #2

    George Jones

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    Multiply out the right side, and then factor out x from all possible terms.
     
  4. Sep 29, 2011 #3

    cepheid

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    Well, it seems like, in this partial fraction decomposition, you can expand the right hand side of your last equation, collect terms, and then solve for A and B.
     
  5. Sep 29, 2011 #4
    [tex]2x = A(3x+1) + B(3+x)[/tex]
    [tex]2x = x(3A+B) + A +3B[/tex]

    ?
     
  6. Sep 29, 2011 #5

    George Jones

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    left = right.

    How many x's on the left? On the right?

    What is the constant on the left? On the right?
     
  7. Sep 29, 2011 #6
    So,

    The constant is A + 3B

    The other equation is 2=(3A+B) ?

    I'm guessing I system of equation these guys to find A and B now? What does the constant equal? 0?
     
  8. Sep 29, 2011 #7
    That makes this I believe:

    [tex]\frac{2x}{(3+x)(3x+1)} = \frac{\frac{3}{4}}{3+x} + \frac{\frac{-1}{4}}{3x+1}[/tex]

    Does that look correct? I can integrate those.
     
  9. Sep 29, 2011 #8

    cepheid

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    Yeah, it seems like you've got it. You can always check your answer by doing the reverse (combine the two terms on the right hand side into one fraction with a common denominator and check that the numerator simplifies to 2x).
     
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