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Finding the antiderivative for fractions?

  1. Dec 4, 2005 #1
    Finding the antiderivative for fractions???


    I get antiderivatives and the idea behind them. But I still don't really comprehend how to apply it towards a fraction.

    We know that [itex]\int (\frac{1}{x}) dx = ln|x|[/itex]

    So would the antiderivative of [itex]\int (\frac{4}{x}) dx = 4(ln x)[/itex] ?

    But here is a fraction I just am not sure where to start?
    [itex]\int (\frac{4}{3x^2}) dx[/itex]

    Any help would be greatly appreciated. :smile:
    Last edited: Dec 4, 2005
  2. jcsd
  3. Dec 4, 2005 #2
    Your first antiderivative is correct. You should probably add an arbitrary constant unless the question asks for an antiderivative.

    For the second one you can factor out constants and then use the power rule.

    \int {\frac{4}{{3x^2 }}} dx = \frac{4}{3}\int {\frac{1}{{x^2 }}} dx = \frac{4}{3}\int {x^{ - 2} dx}
  4. Dec 4, 2005 #3
    Thanks Benny. I'm glad you mentioned the arbitrary constant. I forgot about that. I just get easily confused when it comes to taking the antiderivative of fractions. So lets say I had a problem like:

    [itex]\int \frac{4x}{x}dx[/itex]

    Could I say that the x's cancel so I'd be left with [itex]\int 4dx[/itex] which would equal [itex]4x[/itex]? This however doesn't make sense. Because the derivative of [itex]4x[/itex] would be 4.

    So since I'm seeing that my first choice answer wouldn't be correct. I don't know what to do. LOL Could I take out the 4x?
  5. Dec 4, 2005 #4


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    There is no need to take out the 4x. You are corrrect that the x's will cancel out leaving you with the integral of 4dx. This will give you 4x + c unless of course it integral is bounded.

    The derivative of 4*x is 4. So it is true that what you said is all equal.

    what you are probably not seeing is dv = 4dx. and so you take the integral of both sides and that equals v = 4x.

    the derivative however would be dv/dx = 4x = 4.

    Does this help?
  6. Dec 4, 2005 #5
    Somewhat. I do have more questions, but they will have to be asked/answered possibly tomorrow. Thanks to the both of you for helping me out. :smile:
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