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Homework Help: Remembering how to integrate

  1. Sep 12, 2006 #1


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    It has been about 2 years since i last did calculus but im trying to get back into it so im ready for college / dont kill myself because of boredom

    I am having difficulty finding integrals of the form
    [tex] \int{\frac{x^2}{x+a}dx} [/tex]

    This integral inparticular is:

    I couldnt find a [itex]u[/itex] or a [itex] du [/itex] that would work

    I then tried to decompose the fraction to [itex] \frac{x^2}{2x+2} = \frac{A}{2} + \frac{B}{x+1}[/itex] but could't solve for A, which makes sense, so that wont work.

    Next i tied integration by parts
    [tex] \int{vdu} = uv - \int{udv} [/tex]

    where i used [itex] u=.5 ln|2x+2|[/itex],[itex]du=(2x+2)^{-1}[/itex],[tex]v=x^2[/itex],[itex]dv=2x[/itex].


    [tex]\int{\frac{x^2}{2x+2}dx} = .5x^2 ln |2x+2| - \int{ln |2x+2|xdx} [/tex]

    for the second integral I have:
    [itex] u=.5x^2[/itex],[itex]du=xdx[/itex],[tex]v=ln |2x+2| [/itex],[itex]dv=2(2x+2)^{-1}[/itex].


    [tex]\int{ x ln|2x+2| dx} = .5x^2 ln |2x+2| - \int{\frac{x^2}{2x+2}dx} [/tex]

    and finally both sides cancel out:

    [tex]\int{\frac{x^2}{2x+2}dx} = .5x^2 ln |2x+2| - .5x^2 ln |2x+2| +\int{\frac{x^2}{2x+2}dx} [/tex]

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

    Ive checked the trig rules as well as the inverse trig rules and no other rules match. Does an integral for this function not exsist or am I missing something?
    Last edited: Sep 12, 2006
  2. jcsd
  3. Sep 12, 2006 #2


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    Homework Helper

    When you are integrating a fraction, of which teh degree in the numerator is greater than or equal to the degree of the denominator, one should use Polynomial Long Division first, then use Partial Fraction.
    I'll give you an example:
    Say, you want to integrate:
    [tex]\int \frac{x ^ 2}{x + 1} dx[/tex]
    The degree in the numerator is 2, which is greater than the degree of the denominator 1. So we should use Polynomial Long Division first to get:
    [tex]\int \frac{x ^ 2}{x + 1} dx = \int \left( x - 1 + \frac{1}{x + 1} \right) dx = \int (x - 1) dx + \int \frac{dx}{x + 1}[/tex]
    From here, we can use the substitution u = x + 1 to solve the second integral. So the result is:
    [tex]\int \frac{x ^ 2}{x + 1} dx = \int (x - 1) dx + \int \frac{dx}{x + 1} = \frac{x ^ 2}{2} - x + \ln |x + 1| + C[/tex].
    Ok, can you get this? :)
    Can you go from here? :)
    Last edited: Sep 12, 2006
  4. Sep 12, 2006 #3


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    ?? Partial fractions requires "proper" fractions- that is, that the numerator has lower dimension than the denominator. In this case, you need to divide first.
    [tex]\frac{x^2}{2x+2}= \frac{1}{2}x- \frac{1}{2}- \frac{\frac{1}{2}}{x+1}}[/tex]

    That should be easy to integrate.
  5. Sep 12, 2006 #4


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    polynomial long division rings a bell, i completely forgout about it. im good from there
  6. Sep 12, 2006 #5


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    i didnt know the stipulation about decomposing the fraction, but that makes sense. thanks!
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