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Integral help

  1. Sep 6, 2013 #1
    Integral help!!

    1. The problem statement, all variables and given/known data
    Evaluate the following inde nite integral,
    ∫[itex]\frac{x^3-5x}{x^3-2x^2+4x-8}[/itex] dx


    2. Relevant equations
    I think I am right until i get to equating the equations when I do it the method im taught in class I get c=0 and I dont think that is right. However when i put it into a system of equations with the coefficients i get fractions.

    I would like to know what to do next. Thanks



    3. The attempt at a solution

    First the degree of numerator is same as the denomenator. So i do long division and end up with

    ∫1+ [itex]\frac{2x^2-9x+8}{x^3-2x^2+4x-8}[/itex]

    Now decompose the fraction part into:

    [itex]\frac{2x^2-9x+8}{(x^2+4)(x-2)}[/itex] =[itex]\frac{Ax+B}{(x^2+4)}[/itex]+[itex]\frac{C}{(x-2)}[/itex]

    Then get a common denomenator for RHS
    = [itex]\frac{(Ax+B)(x-2)+C(x^2+4)}{(x^2+4)(x-2)}[/itex]

    From here i tried if x=2 to cancel out the (Ax+B) term which results in C=0
    this is where i believe an error has occured.

    My other thinking was can I expand then factor out the x^2 and x and constant leaving the A and B and C in whatever they factor into and then do a system of equations.
    My result this way is
    A= 9/4, B =-9/2, C=-1/4

    Kind regards
     
  2. jcsd
  3. Sep 6, 2013 #2

    CAF123

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    Gold Member

    Those values of A,B and C are correct. However, in your initial attempt, you should not get C=0.
     
  4. Sep 6, 2013 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    So the two numerators are equal: [itex]2x^2- 9x+ 8= (Ax+B)(x- 2)+ C(x^2+ 4)[/itex]

    Taking x= 2 gives 8- 18+ 8= 8C which gives C= -1/4, as you have below, not 0. Did you forget the left side of the equation?

    If, in [itex]2x^2- 9x+ 8= (Ax+B)(x- 2)+ C(x^2+ 4)[/itex], you take x= 0 (just because it is an easy number) you get [itex]8= (0+ B)(-2)+ C(4)[/itex] or [itex]-2B= -8- (-1/4)(4)= -9[/itex] so that C= -9/2, again as you say. Finally, taking x= 1 (again just because it is an easy number) gives [itex]2- 9+ 8= (A+ B)(-1)+ C(5)[/itex] or [itex]1= -A+ 9/2- 5/4[/itex] so that [itex]A= 9/2- 1- 5/4= 18/4- 4/4- 5/4= 9/4[/itex], again, exactly what you have.

    Great! Now all you need to do is the integral:
    [tex]\int dx+ \dfrac{9}{4}\int \dfrac{xdx}{x^2+ 4}- \dfrac{9}{2}\int \dfrac{dx}{x^2+ 4}- \dfrac{1}{4}\int \dfrac{dx}{x- 2}[/tex]
     
  5. Sep 6, 2013 #4
    Thanks for the reply, looking again at how i got C=0 was most likely due to it being very late at night :).
    But yes I find it easier solving for A,B,C in the system of equations. I see now that my thinking was on the right track and it was just the small error that through me off. Thanks for the help.
     
  6. Sep 6, 2013 #5
    Is my final answer correct?

    x + [itex]\frac{9}{8}[/itex] log | x2+4 | - [itex]\frac{9}{4}[/itex] arctan([itex]\frac{x}{2}[/itex]) - [itex]\frac{1}{4}[/itex] log | x-2 | + C
     
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