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Stuck with an integral

  1. Jan 20, 2013 #1
    1. The problem statement, all variables and given/known data
    Integrate


    2. Relevant equations

    I started with (2x^2+3)/(x^3-2x^2+3x) dx


    3. The attempt at a solution

    Got dx/x + (x+2)/(x^2-2x+3) dx using the A/x + (Bx+C)/(x^2-2x+3) method.

    But I can't seem to solve (x+2)/(x^2-2x+3) dx

    Possible solutions are welcome.
     
    Last edited: Jan 20, 2013
  2. jcsd
  3. Jan 20, 2013 #2

    LCKurtz

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    Welcome to PF. You won't get solutions here, but you may get hints. Try this:$$
    \frac {x+2}{x^2-2x+3} = \frac{(x-1)+3}{(x-1)^2+2}$$and see if that gives you any ideas.
     
  4. Jan 20, 2013 #3
    I thought of that but the +2 at the bottom ruins it, because splitting up x-1 and 3 into 2 integrals won't help, as I cant cancel anything out
     
  5. Jan 20, 2013 #4

    LCKurtz

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    But you can break it up into two fractions and use different methods on them.
     
  6. Jan 20, 2013 #5

    lurflurf

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    $$2x^2+3=(x^2-x)+(3x)+(x^2-2x+3)=\frac{x}{2}\left(\frac{x^3-2x^2+3x}{x}\right)^\prime+3\frac{x^3-2x^2+3x}{x^2-2x+3}+\frac{x^3-2x^2+3x}{x}$$
     
  7. Jan 20, 2013 #6
    Got it solved. Thank you for the help.
     
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